chap11

Chapter 11 Organizer: Stoichiometry

Section Objectives National Standards

State/Local

StandardsResources to Assess Mastery

Section 11.11. Describe the types of relationships

indicated by a balanced chemical equation.

2. State the mole ratios from a balanced chemical equation.

UCP.1, UCP.3; A.1; B.3; E.1

Entry-Level Assessment Focus Transparency 40Progress Monitoring Formative Assessment, p. 371 Reading Check, pp. 369, 371 Section Assessment, p. 372

Section 11.21. List the sequence of steps used in

solving stoichiometric problems.2. Solve stoichiometric problems.

UCP.1, UCP.3; A.1; B.3; E.1

Entry-Level Assessment Focus Transparency 41Progress Monitoring Formative Assessment, p. 374 Section Assessment, p. 378

Section 11.31. Identify the limiting reactant in a

chemical equation.2. Identify the excess reactant, and

calculate the amount remaining after the reaction is complete.

3. Calculate the mass of a product when the amounts of more than one reactant are given.

UCP.1, UCP.3; B.3; E.1, E.2; F.1

Entry-Level Assessment Focus Transparency 42Progress Monitoring Formative Assessment, pp. 381, 383, 384 Reading Check, p. 380 Section Assessment, p. 384

Section 11.41. Calculate the theoretical yield of

a chemical reaction from data.2. Determine the percent yield for

a chemical reaction.

UCP.1, UCP.3; A.1, A.2; B.3, B.6; E.1, E.2; F.4, F.5, F.6; G.1

Entry-Level Assessment Focus Transparency 43Progress Monitoring Formative Assessment, p. 388 Section Assessment, p. 388Summative Assessment Chapter Assessment, p. 392 ExamView® Assessment Suite CD-ROM

Mass relationships in chemical reactions confirm the law of conservation of mass.

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BL Below Level OL On Level AL Advanced Learners EL English Learners COOP LEARN Cooperative Learning

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Leveled Resources Lab Materials Additional Print andTechnology Resources

Science Notebook 11.1 OLFAST FILE Chapter Resources: Study Guide, p. 72 OLTransparencies: Section Focus Transparency 40 BL EL Math Skills Transparency 15 OL EL

Launch Lab, p. 367: 10-mL graduated cylinder, 0.01M potassium permanganate, 100-mL beaker, 0.01M sodium hydrogen sulfite solution, stirring rod 15 min

Technology: ExamView® Assessment Suite CD-ROM StudentWorks™ Plus DVD-ROM TeacherWorks™ Plus DVD-ROM Virtual Labs CD-ROM Video Labs DVD Interactive Classroom DVD-ROM LabManager™ CD-ROM

Assessment: Performance Assessment in the Science Classroom Challenge Problems AL Supplemental Problems BL OL Chapter Test (Scaffolded)

FAST FILE Resources: Section Focus Transparency Masters Math Skills Transparency Masters and Worksheets Teaching Transparency Masters and Worksheets

Additional Resources: Solving Problems: A Chemistry Handbook Cooperative Learning in the Science Classroom Lab and Safety Skills in the Science Classroom glencoe.com

Lab Resources: Laboratory Manual OL CBL Laboratory Manual OL Small-Scale Laboratory Manual OL Forensics Laboratory Manual OL

Science Notebook 11.2 OLFAST FILE Chapter Resources: MiniLab Worksheet, p. 58 OL ChemLab Worksheet, p. 60 OL Study Guide, p. 74 OLTransparencies: Section Focus Transparency 41 BL EL

Teaching Transparency 35 OL EL Math Skills Transparencies 16, 17 OL EL

MiniLab, p. 378: balance, crucible, sodium hydrogen carbonate (NaHCO3), ring stand, ring, clay triangle, Bunsen burner, crucible tongs 25 minChemLab, p. 390: copper(II) sulfate pentahydrate, iron metal filings (20 mesh), distilled water, hot plate, beaker tongs, balance, stirring rod, 150-mL beaker, 400-mL beaker, 100-mL graduated cylinder, weighing paper 45 min

Science Notebook 11.3 OLFAST FILE Chapter Resources: Study Guide, p. 76 OLTransparencies: Section Focus Transparency 42 BL EL Teaching Transparency 36 OL EL

Science Notebook 11.4 OLFAST FILE Chapter Resources: Study Guide, p. 77 OLTransparencies: Section Focus Transparency 43 BL EL

Suggested PacingPeriod Section 11.1 Section 11.2 Section 11.3 Section 11.4 Assessment

Single 1 2 1 1 1

Block 0.5 1 0.5 0.5 0.5

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BIG IdeaReactive Substances To intro-duce this chapter’s Big Idea, ask four students to each add 5.0 mL of water to an empty film canister. Tell them to divide a sodium bicarbonate tablet into four unequal pieces—small, medium, medium-large, and large. Have them place each piece of the tablet on the lid of a film canister. When you say so, students should snap the lid onto the canister, invert the canister on the table, and back away from the table. Film canisters will pop and be launched into the air. Ask students which canister had the biggest pop. The canister with the largest piece of the tablet will likely have the biggest pop. Ask students if the size of the sodium bicarbonate in the canister determined how high the canister traveled. Results will likely correlate to the size of the tablet.

Tie to Previous KnowledgeHave students review the following concepts before studying this chapter.Chapter 2: scientific notation, significant figures, dimensional analysisChapter 4: average atomic massChapter 10: mole conversions

Use the PhotoConservation Write the balanced basic equation for photosynthesis on the board. Remind students of the law of the conservation of mass, which was introduced in Chapter 3, Section 2. Ask them to verify that mass is conserved during photosyn-thesis. Students should verify that the total number of each type of atom, and therefore the mass, is identical on both sides of the equation.

This DVD-ROM is an editable Microsoft® PowerPoint® presentation that includes:• a premade presentation for every chapter• additional diagnostic, formative, chapter, and Standardized Test Practice questions• animations• image bank• transparencies• links to glencoe.com

Interactive Classroom

Stoichiometry

BIG Idea Mass relationships in chemical reactions confirm the law of conservation of mass.

11.1 Defining StoichiometryMAIN Idea The amount of each reactant present at the start of a chemical reaction determines how much product can form.

11.2 Stoichiometric CalculationsMAIN Idea The solution to every stoichiometric problem requires a balanced chemical equation.

11.3 Limiting Reactants MAIN Idea A chemical reaction stops when one of the reactants is used up.

11.4 Percent Yield MAIN Idea Percent yield is a measure of the efficiency of a chemical reaction.

ChemFacts

• Green plants make their own food through photosynthesis.

• Photosynthesis occurs within structures called chloroplasts in the cells of plants.

• The balanced chemical equation for the photosynthesis is:

6C O 2 + 6 H 2 O → C 6 H 12 O 6 + 6 O 2

• On a summer day, one acre of corn produces enough oxygen (a product of photosynthesis) to meet the respiratory needs of 130 people.

Carbon dioxide and water

Chloroplast

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Start-Up Activities

LAUNCH Lab

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LAUNCH LabLabRUBRIC available at glencoe.com

Purpose Students will observe a chemical reaction in which one of the reactants is completely used up and the other is in excess.

Safety Precautions Approve lab safety forms before work begins. Review the MSDS for potassium permanganate and sodium hydrogen sulfite with students prior to doing the lab. Do not allow students to prepare the solutions required for the lab, as KMn O 4 is a dangerous oxidizer in powder and crystal form. Remind students to wear aprons and goggles.

Disposal The KMn O 4 solution must be processed or saved as chem-ical waste—do not pour the solution down the drain.

Teaching Strategies• Place five test tubes and ten rubber

stoppers on the desk. Place one stopper into each test tube, and ask students which item is in excess.

• See page 48T for preparation of all solutions.

Expected Results Initially, the KMn O 4 solution is dark purple. The solution gradually becomes colorless as the NaHS O 3 solution is added. Expect that 5 mL KMn O 4 solution will require 27 mL NaHS O 3 solution to change it to a clear, colorless solution.

Analysis 1. As colorless sodium hydrogen sulfite was added to

the purple potassium permanganate solution, a color change from purple to colorless was observed.

2. Adding the solution all at once could lead to an error in the volume of NaHSO 3 solution needed to change the purple KM nO 4 solution to a colorless solution. The error could be as large as 5.0 mL.

Inquiry Nothing more will happen because the solution is colorless, meaning that there is no more potassium permanganate to react.

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Start-Up ActivitiesStart-Up Activities

LLAAUUNCH NCH LabLabWhat evidence can you observe that a reaction is taking place?During a chemical reaction, reactants are consumed as new products are formed. Often, there are several telltale signs that a chemical reaction is taking place.

Procedure 1. Read and complete the lab safety form.2. Use a 10-mL graduated cylinder to measure

out 5.0 mL 0.01M potassium permanganate (KMn O 4 ). Add the solution to a 100-mL beaker.

3. Clean and dry the graduated cylinder, and then use it to measure 5.0 mL 0.01M sodium hydrogen sulfite solution (NaHS O 3 ). Slowly add this solution to the beaker while stirring with a stirring rod. Record your observations.

4. Repeat Step 3 until the KMn O 4 solution in the beaker turns colorless. Stop adding the NaHS O 3 solution as soon as you obtain a colorless solution. Record your observations.

Analysis1. Identify the evidence you observed that a chemical

reaction was occurring. 2. Explain why slowly adding the NaHS O 3 solution while

stirring is a better experimental technique than adding 5.0 mL of the solution all at once.

Inquiry Would anything more have happened if you continued to add NaHS O 3 solution to the beaker? Explain.

Steps in Stoichiometric Calculations Make the following Foldable to help you summarize the steps in solving a stoichiometric problem.

Visit glencoe.com to: study the entire chapter online

explore

take Self-Check Quizzes

use the Personal Tutor to work Example Problems step-by-step

access Web Links for more information, projects, and activities

find the Try at Home Lab, Baking Soda Stoichiometry

STEP 1 Fold a sheet of paper in half lengthwise.

STEP 2 Fold in half widthwise and then in half again.

STEP 3 Unfold and cut along the folds of the top flap to make four tabs.

STEP 4 Label the tabs with the steps in stoichiometric calculations.

Use this Foldable with Section 11.2. As you read this section, summarize each step on a tab and include an example of the step.

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Section 11.111.11 Focus

Focus TransparencyBefore presenting the lesson, project Section Focus Transparency 40 and have students answer the accompa-nying questions. BL EL

MAIN IdeaReactants and Products Place a candle on a glass plate and place the plate into a bowl of water. Light the candle, and ask students what will happen if you place a beaker over the candle. The candle will go out when it uses up the oxygen in the beaker. Place the beaker over the candle and have stu-dents watch the water level rise in the beaker. Ask students why the water level increased. In the balanced equation, moles of gaseous reactants are greater than moles of gaseous products, resulting in lower pressure inside the beaker after the candle burns out and cools. Tell students that this chapter (Stoichi-ometry) involves amounts of reac-tants and products. OL

Differentiated Instruction

Below Level Using a molecular model kit, make a model of C H 4 and two O 2 molecules. Ask students to identify the models. Make models of C O 2 and two H 2 O molecules and ask students to identify these models. Tell students that the two sets of models represent the reactants and the products of the combustion of methane. Arrange the reactants and products on opposite sides of a cardboard arrow. Ask students to count the number of each atom before and after the reaction. one carbon, four hydrogen, and four oxygen atoms Remind them that coefficients indicate the number of moles as well as the number of molecules. Have them interpret the equation in moles. 1 mol C H 4 + 2 mol O 2 → 1 mol C O 2 + 2 mol H 2 O BL EL

Quick Demo

Combustion Attach the ends of a rubber hose to a small glass funnel and a gas jet. Fill a petri dish with a soap solution con-taining 10% dish-washing soap and a few drops of glycerol. Dip the funnel into the bubble mix-ture. Attach a small wooden splint to the end of a meter stick and ignite the splint with a match. Turn on the gas jet just enough to slowly produce bub-bles in the funnel. Shake the bubbles loose from the funnel and have a student touch each bubble with the burning splint. The bubbles will ignite. Remind students that you are igniting bubbles of natural gas or meth-ane (C H 4 ). Ask students to write the combustion equation. CH 4 + 2 O 2 → C O 2 + 2H 2 O OL EL

Section 11.111.1

Objectives

◗◗ Describe the types of relationships indicated by a balanced chemical equation.

◗◗ State the mole ratios from a balanced chemical equation.

Review Vocabularyreactant: the starting substance in a chemical reaction

New Vocabularystoichiometrymole ratio

Defining StoichiometryMAIN Idea The amount of each reactant present at the start of a chemical reaction determines how much product can form.

Real-World Reading Link Have you ever watched a candle burning? You might have watched the candle burn out as the last of the wax was used up. Or, maybe you used a candle snuffer to put out the flame. Either way, when the candle stopped burning, the combustion reaction ended.

Particle and Mole RelationshipsIn doing the Launch Lab, were you surprised when the purple color of potassium permanganate disappeared as you added sodium hydrogen sulfite? If you concluded that the potassium permanganate had been used up and the reaction had stopped, you are right. Chemical reactions stop when one of the reactants is used up. When planning the reaction of potassium permanganate and sodium hydrogen sulfite, a chemist might ask, “How many grams of potassium permanganate are needed to react completely with a known mass of sodium hydrogen sulfite?” Or, when analyzing a photosynthesis reaction, you might ask, “How much oxygen and carbon dioxide are needed to form a known mass of sugar.” Stoichiometry is the tool for answering these questions.

Stoichiometry The study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemi-cal reaction is called stoichiometry. Stoichiometry is based on the law of conservation of mass. Recall from Chapter 3 that the law states that matter is neither created nor destroyed in a chemical reaction. In any chemical reaction, the amount of matter present at the end of the reac-tion is the same as the amount of matter present at the beginning. Therefore, the mass of the reactants equals the mass of the products. Note the reaction of powdered iron (Fe) with oxygen ( O 2 ) shown in Figure 11.1. Although iron reacts with oxygen to form a new com-pound, iron(III) oxide (F e 2 O 3 ), the total mass is unchanged.

■ Figure 11.1 The balanced chemi-cal equation for this reaction between iron and oxygen provides the relation-ships between amounts of reactants and products.

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Visual LearningTable 11.1 Ask students to use the table as a guide for interpreting the equation for the combustion of methane. Make a drawing of the reaction to help students. OL EL

Interactive Table Students can interact with the table at glencoe.com.

Reading Check The coefficients in a balanced chemical equation represent the number of representative particles as well as the numbers of moles of particles. Though they do not directly indicate anything about the masses of the reactants or particles, these masses can be derived from the coefficients by converting the known mole quantities to mass.

Quick Demo

Quantities of Reactants Take four 125 mL flasks and four balloons. To each flask, add 10 mL of 2.0M acetic acid. Mass out 0.5, 1.0, 2.5, and 5.0 g sam-ples of baking soda on weighing paper. Ask students what will happen when the baking soda is added to the vinegar. A reaction will take place that releases a gas. Ask students if all flasks will produce the same amount of carbon dioxide. Answers will vary. Place the baking soda into the acetic acid in the flasks and quickly cover the top of each flask with the balloon. Ask students why all of the balloons were not the same size. Each reaction produced a different amount of carbon dioxide, which depended on the amount of acetic acid and baking soda present. BL EL

Chemistry JournalInterpret Chemical Equations Have students interpret the following equations in terms of particles, moles, and mass.2KC lO 3 → 2KCl + 3 O 2 Fe 2 O 3 + 3 H 2 → 2Fe + 3 H 2 OHave them demonstrate in their journals that the law of conservation of mass is observed. OL

11.1

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VOCABULARYWORD ORIGIN

Stoichiometrycomes from the Greek words stoikheion, which means element, and metron, which means to measure

Table 11.1

Relationships Derived from a Balanced Chemical Equation

4Fe(s) + 3 O 2 (g) → 2F e 2 O 3 (s)

iron + oxygen → iron(III) oxide

4 atoms Fe + 3 molecules O 2 → 2 formula units F e 2 O 3

4 mol Fe + 3 mol O 2 → 2 mol F e 2 O 3

223.4 g Fe + 96.00 g O 2 → 319.4 g F e 2 O 3

319.4 g reactants → 319.4 g products

Interactive Table Explore balanced chemical equations at glencoe.com.

The balanced chemical equation for the chemical reaction shown in Figure 11.1 is as follows.

4Fe(s) + 3 O 2 (g) → 2F e 2 O 3 (s)

You can interpret this equation in terms of representative particles by saying that four atoms of iron react with three molecules of oxygen to produce two formula units of iron(III) oxide. Remember that coeffi-cients in an equation represent not only numbers of individual particles but also numbers of moles of particles. Therefore, you can also say that four moles of iron react with three moles of oxygen to produce two moles of iron(III) oxide.

The chemical equation does not directly tell you anything about the masses of the reactants and products. However, by converting the known mole quantities to mass, the mass relationships become obvious. Recall that moles are converted to mass by multiplying by the molar mass. The masses of the reactants are as follows.

4 mol Fe × 55.85 g Fe

_ 1 mol Fe = 223.4 g Fe

3 mol O 2 × 32.00 g O 2

_ 1 mol O 2 = 96.00 g O 2

The total mass of the reactants is: (223.4 g + 96.00 g) = 319.4 g

Similarly, the mass of the product is calculated as follows:

2 mol F e 2 O 3 × 159.7 g F e 2 O 3

__ 1 mol F e 2 O 3 = 319.4 g

Note that the mass of the reactants equals the mass of the product.

mass of reactants = mass of products 319.4 g = 319.4 g

As predicted by the law of conservation of mass, the total mass of the reactants equals the mass of the product. The relationships that can be determined from a balanced chemical equation are summarized in Table 11.1.

Reading Check List the types of relationships that can be derived from the coefficients in a balanced chemical equation.

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Section 11.1 • Defining Stoichiometry 369


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IN-CLASS Example

Question Geologists test for calcium carbonate in rocks using a hydrochloric acid solution. Bubbling indicates the release of carbon diox-ide from the limestone according to the following reaction: CaC O 3 + 2HCl → CaC l 2 + H 2 O + C O 2 Interpret the equation for this reaction in terms of representative particles, moles, and mass. Show that the law of conservation of mass is observed.

Answer 1 formula unit of CaC O 3 + 2 molecules of HCl → 1 formula unit of C aCl 2 + 1 molecule o f H 2 O + 1 molecule o f CO 2 ;1 mol CaCO 3 + 2 mol HCl → 1 mol CaC l 2 + 1 mol H 2 O + 1 mol CO 2 ;100 g CaC O 3 + 75.0 g HCl → 113 g CaC l 2 + 18.0 g H 2 O + 44.0 g175 g reactants = 175 g products. Mass is conserved.

Chemistry ProjectStoichiometry in Space Have students research the role of lithium hydroxide in the removal of the carbon dioxide exhaled by astronauts. Ask them to discuss why the carbon dioxide must be removed, and how stoichiometric calculations are used to minimize the weight of lithium hydroxide carried into space. Ask advanced students to perform a rough calculation of the amount of carbon dioxide exhaled by a person in a day. OL AL

Differentiated InstructionHearing Impaired Have students view a video clip of a shuttle launch. Show on the board that the fuel for the launch is a mixture of hydrogen and oxygen. Have students write a paragraph describing what factors engineers have to consider when using these fuels. Paragraphs should include such factors as how much hydrogen is present and how much oxygen is needed. OL

PRACTICE ProblemsHave students refer to p. 998 for complete solutions to odd-numbered problems. The complete solutions for all problems can be found in the Solutions Manual.

1. a. 34.062 g reactants = 34.062 g products

b. 92.566 g reactants = 92.566 g products

c. 80.608 g reactants = 80.608 g products

2. a. 2Na(s) + 2 H 2 O(l) → 2NaOH(aq) +1H 2 (g)82.01 g reactants = 82.01 g products

b. 4Zn(s) + 10HN O 3 (aq) →4Zn(N O 3 ) 2 (aq) + 1N 2 O( g) + 5H 2 O(l)891.68 g reactants = 891.68 g products

EXAMPLE Problem 11.1

Interpreting Chemical Equations The combustion of propane ( C 3 H 8 ) provides energy for heating homes, cooking food, and soldering metal parts. Interpret the equation for the combustion of propane in terms of representative particles, moles, and mass. Show that the law of conservation of mass is observed.

1 Analyze the ProblemThe coefficients in the balanced chemical equation shown below represent both moles and representative particles, in this case molecules. Therefore, the equation can be interpreted in terms of molecules and moles. The law of conservation of mass will be verified if the masses of the reactants and products are equal.

KnownC3H8(g) + 5 O2(g) → 3C O2(g) + 4 H2O(g)

UnknownEquation interpreted in terms of molecules = ?Equation interpreted in terms of moles = ?Equation interpreted in terms of mass = ?

2 Solve for the UnknownThe coefficients in the chemical equation indicate the number of molecules.

1 molecule C3H8 + 5 molecules O2→ 3 molecules CO2 + 4 molecules H2O

The coefficients in the chemical equation also indicate the number of moles.

1 mol C3H8 + 5 mol O2→ 3 mol CO2 + 4 mol H2O

To verify that mass is conserved, first convert moles of reactant and product to mass by multiplying by a conversion factor—the molar mass—that relates grams to moles.

moles of reactant or product ×grams reactant or product___1 mol reactant or product

= grams of reactant or product

1 mol C3H8 ×44.09 g C3H8__1 mol C3H8

= 44.09 g C3H8 Calculate the mass of the reactant C3H8.

5 mol O2 ×32.00 g O2_1 mol O2

= 160.0 g O2 Calculate the mass of the reactant O2.

3 mol C O2 ×44.01 g C O2_1 mol C O2

= 132.0 g C O2 Calculate the mass of the product CO2.

4 mol H2O ×18.02 g H2O_1 mol H2O

= 72.08 g H2O Calculate the mass of the product H2O

44.09 g C3H8 + 160.0 g O2 = 204.1 g reactants Add the masses of the reactants.

132.0 g C O2 + 72.08 g H2O = 204.1 g products Add the masses of the products.

204.1 g reactants = 204.1 g products The law of conservation of mass is observed.

3 Evaluate the Answer The sums of the reactants and the products are correctly stated to the first decimal place because each mass is accurate to the first decimal place. The mass of reactants equals the mass of products, as predicted by the law of conservation of mass.

Roundingpage 952

Math Handbook




AssessmentAssessmentSkill Have each student write

a balanced equation on a sheet of paper and give it to another student, who will write the mole ratios found in the equation. OL

3 AssessCheck for UnderstandingPlace a small piece of calcium into a test tube containing 3.0M HCl. In a second test tube, collect the hydro-gen gas. Demonstrate the presence of hydrogen in the test tube using a burning splint. Note the characteris-tic pop. On the board, write the word equation for the reaction: calcium plus hydrochloric acid produces cal-cium chloride and hydrogen. Ask students to write the balanced equa-tion. Ca + 2HCl → Ca Cl 2 + H 2 Have stu-dents write at least two mole ratios for the equation. OL

ReteachPour 10 mL of 0.10M Ba(O H) 2 into a test tube. Add 10 mL of 0.10Msulfuric acid. Ask students to write the equation for the reaction that occurs and all the possible mole ratios. Ba(O H) 2 + H 2 S O 4 → Ba SO 4 + 2 H 2 OOL

ExtensionAsk students to explain how mole ratios derived from a balanced chemical equation can be used to relate the masses of reactants and products in a reaction. Because molar quantities are related to mass by Avogadro’s number, mole ratios can be converted to mass ratios. OL

Reading Check A chemical reaction’s mole ratios are derived from the relationships between coefficients in a bal-anced chemical equation. A mole ratio is a ratio between the numbers of moles of any two substances in the equation.

Virtual Labs CD-ROMChemistry: Matter and ChangeExploration: Predicting Mass of Products

GLENCOE TechnologyVirtual Lab

CD-ROM How Much Oxygen is Available?

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Mole ratios You have read that the coefficients in a chemical equa-tion indicate the relationships between moles of reactants and products. You can use the relationships between coefficients to derive conversion factors called mole ratios. A mole ratio is a ratio between the numbers of moles of any two of the substances in a balanced chemical equation. For example, consider the reaction shown in Figure 11.2. In this reaction, potassium (K) reacts with bromine (B r 2 ) to form potassium bromide (KBr). The product of the reaction, the ionic salt potassium bromide, is prescribed by veterinarians as an antiepileptic medication for dogs and cats.

2K(s) +B r 2 (l) → 2KBr(s)

What mole ratios can be written for this reaction? Starting with the reactant potassium, you can write a mole ratio that relates the moles of potassium to each of the other two substances in the equation. Thus, one mole ratio relates the moles of potassium used to the moles of bro-mine used. The other mole ratio relates the moles of potassium used to the moles of potassium bromide formed.

2 mol K _ 1 mol B r 2 and 2 mol K _ 2 mol KBr

Two other mole ratios show how the moles of bromine relate to the moles of the other two substances in the equation—potassium and potassium bromide.

1 mol B r 2 _ 2 mol K and 1 mol B r 2 _ 2 mol KBr

Similarly, two ratios relate the moles of potassium bromide to the moles of potassium and bromine.

2 mol KBr _ 2 mol K and 2 mol KBr _ 1 mol B r 2

These six ratios define all the mole relationships in this equation. Each of the three substances in the equation forms a ratio with the two other substances.

Reading Check Identify the source from which a chemical reaction’s mole ratios are derived.

■ Figure 11.2 Potassium metal and liquid bromine react vigorously to form the ionic compound potassium bromide. Bromine is one of the two elements that are liquids at room temperature (mercury is the other). Potassium is a highly reactive metal.

PRACTICE Problems Extra Practice Pages 982–983 and glencoe.com

1. Interpret the following balanced chemical equations in terms of particles, moles, and mass. Show that the law of conservation of mass is observed.

a. N2(g) + 3 H2(g) → 2N H3(g)

b. HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

c. 2Mg(s) + O2(g) → 2MgO(s)

2. Challenge For each of the following, balance the chemical equation; interpret the equation in terms of particles, moles, and mass; and show that the law of conservation of mass is observed.

a. ___Na(s) + ___ H2O(l) → ___NaOH(aq) + ___ H2(g)

b. ___Zn(s) + ___HN O3(aq)→ ___Zn(N O3)2(aq) + ___ N2O(g) + ___ H2O(l)

Personal Tutor For an online tutorial on ratios, visit glencoe.com.

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Section 11.1 • Defining Stoichiometry 371


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Self-Check Quiz glencoe.com


3. a. 4 mol Al __ 3 mol O 2

3 mol O 2

__ 2 mol Al 2 O 3

2 mo l Al 2 O 3

__ 4 mol Al

3 mol O 2

__ 4 mol Al

2 mol Al 2 O 3

__ 3 m ol O 2

4 mol Al __ 2 m ol Al 2 O 3

b. 3 mol Fe __ 4 mo l H 2 O

3 mol Fe __ 4 mol H 2

3 mol Fe __ 1 m ol F e 3 O 4

4 mol H 2 O

__ 3 mol Fe

4 m ol H 2

__ 3 mol Fe

1 mol Fe 3 O 4

__ 3 mol Fe

1 mol Fe 3 O 4 __4 mol H 2

1 mol Fe 3 O 4

__ 4 mol H 2 O

4 m ol H 2 O

__ 4 mol H 2

4 m ol H 2

__ 1 mo l Fe 3 O 4

4 mo l H 2 O

__ 1 mo l Fe 3 O 4

4 m ol H 2

__ 4 mol H 2 O

c. 2 mol HgO

__ 2 mol Hg

1 mol O 2

__ 2 mol Hg

1 mol O 2

__ 2 mol HgO

2 mol Hg

__ 2 mol HgO

2 mol Hg

__ 1 mol O 2

2 mol HgO

__ 1 mol O 2

4. a. ZnO + 2HCl → ZnCl 2 + H 2 O

1 mol ZnO__2 mol HCl

1 mol ZnO__1 mol ZnC l 2

1 mol ZnO__1 mol H 2 O

2 mol HCl __ 1 mol ZnO

2 mol HCl __ 1 mol Z nCl 2

2 mol HCl __ 1 m ol H 2 O

1 mol Z nCl 2

__ 1 mol ZnO

1 mol Zn Cl 2

__ 2 mol HCl

1 mol ZnC l 2

__ 1 mol H 2 O

1 mol H 2 O

__ 1 mol ZnO

1 mo l H 2 O

__ 2 mol HCl

1 mol H 2 O

__ 1 mol Z nCl 2

b. 2 C 4 H 10 + 13O 2 → 8C O 2 + 1 0H 2 O2 mol C 4 H 10 __13 mol O 2

2 mo l C 4 H 10 __8 mol CO 2

2 mol C 4 H 10 __10 mol H 2 O

13 mol O 2

__ 2 mo l C 4 H 10

8 mol CO 2

__ 2 mo l C 4 H 10

10 mol H 2 O

__ 2 mo l C 4 H 10

10 mol H 2 O

__ 13 mol O 2

10 mo l H 2 O

__ 8 mo l CO 2

8 mol C O 2

__ 13 mol O 2

13 mol O 2

__ 10 mol H 2 O

8 mol C O 2

__ 10 mo l H 2 O

13 m ol O 2

__ 8 mo l CO 2

Section 11.111.1 Assessment 5. The coefficients in the balanced equation indicate the molar relationship

between each pair of reactants and products. 6. (3)(2) = 6 ratios 7. particles (atoms, molecules, formula units), moles, and mass 8. xA/yB and xA/zAB, yB/xA and yB/zAB, zAB/xA and zAB/yB

9. 2 H 2 O 2 → 2 H 2 O + O 2 2 mo l H 2 O 2 /2 mol H 2 O, 2 m ol H 2 O 2 /1 mo l O 2 , 2 mol H 2 O/2 mol H 2 O 2 , 2 mol H 2 O/1 mo l O 2 , 1 mol O 2 /2 m ol H 2 O 2 , 1 m ol O 2 /2 mol H 2 O 10. 2H 2 /O 2 an d 2H 2 /2H 2 O, O 2 /2H 2 and O 2 /2H 2 O, 2H 2 O/2 H 2 and 2 H 2 O/O 2 Student sketches should show six hydrogen molecules reacting with three

oxygen molecules to form six water molecules.

Section 11.111.1 AssessmentSection Summary◗ ◗ Balanced chemical equations can be

interpreted in terms of moles, mass, and representative particles (atoms, molecules, formula units).

◗ ◗ The law of conservation of mass applies to all chemical reactions.

◗ ◗ Mole ratios are derived from the coef-ficients of a balanced chemical equa-tion. Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction.

5. MAIN Idea Compare the mass of the reactants and the mass of the products in a chemical reaction, and explain how these masses are related.

6. State how many mole ratios can be written for a chemical reaction involving three substances.

7. Categorize the ways in which a balanced chemical equation can be interpreted.

8. Apply The general form of a chemical reaction is xA + y B → zAB. In the equa-tion, A and B are elements, and x, y, and z are coefficients. State the mole ratios for this reaction.

9. Apply Hydrogen peroxide ( H 2 O 2 ) decomposes to produce water and oxygen. Write a balanced chemical equation for this reaction, and determine the possible mole ratios.

10. Model Write the mole ratios for the reaction of hydrogen gas and oxygen gas, 2 H 2 (g) + O 2 (g) → 2 H 2 O. Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Show the water molecules produced.

The decomposition of potassium chlorate (KCl O 3 ) is sometimes used to obtain small amounts of oxygen in the laboratory.

2KCl O 3 (s) → 2KCl(s) + 3 O 2 (g)

The mole ratios that can be written for this reaction are as follows.

2 mol KCl O 3 _ 2 mol KCl and 2 mol KCl O 3

_ 3 mol O 2

2 mol KCl _ 2 mol KCl O 3 and 2 mol KCl _ 3 mol O 2

3 mol O 2 _ 2 mol KCl O 3 and 3 mol O 2 _ 2 mol KCl

Note that the number of mole ratios you can write for a chemical reaction involving a total of n substances is (n)(n–1). Thus, for reactions involving four and five substances, you can write 12 and 20 moles ratios, respectively. Four substances: (4)(3) = 12 mole ratios Five substances: (5)(4) = 20 mole ratios

VOCABULARYACADEMIC VOCABULARY

Deriveto obtain from a specified sourceThe researcher was able to derive the meaning of the illustration from ancient texts.

PRACTICE Problems Extra Practice 83 and glencoe.com

3. Determine all possible mole ratios for the following balanced chemical equations.

a. 4Al(s) + 3 O2(g) → 2A l2O3(s)

b. 3Fe(s) + 4 H2O(l) → F e3O4(s) + 4 H2(g)

c. 2HgO(s)→ 2Hg(l) + O2(g)

4. Challenge Balance the following equations, and determine the possible mole ratios.

a. ZnO(s) + HCl(aq) → ZnC l2(aq) + H2O(l)

b. butane ( C4H10) + oxygen → carbon dioxide + water

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MAIN IdeaAdjust for Quantitative Differences A recipe for a banana split calls for four ingredients, each in a particular quantity: one banana, two scoops of ice cream, 50 mL of chocolate syrup and 50 mL of strawberry syrup. Write the ingredients on the board in the form of a chemical equation:__banana + __ice cream + __

chocolate syrup + __strawberry syrup → __banana split. Then, “balance” the equation by giving the quantities: banana + 2 ice cream + 50 chocolate syrup + 50 strawberry syrup → 1 banana split.Ask students what is needed to make five banana splits following the same recipe. 5 bananas, 10 scoops of ice cream, 250 mL of chocolate syrup, and 250 mL of strawberry syrup. BL OL

2 TeachConcept DevelopmentUsing Moles Point out to students that moles are always involved when solving stoichiometry problems. Remind them that the mole ratio is needed to convert from one sub-stance in the balanced equation to another substance. If they are in doubt about how to proceed in solving a problem, remind them to “go to moles first.” OL

RUBRIC available at glencoe.com

Chemistry ProjectSoap Making Soap making is a multistep process involving several chemical reac-tions. Have students research the process, identify the compounds and substances involved, and summarize the basic reac-tions. In particular, have students write a balanced chemical equation for a reaction used to produce lye. From the equation, ask them to write all of the mole ratios. AL

Section 11.211.2

Objectives

◗◗ List the sequence of steps used in solving stoichiometric problems.

◗◗ Solve stoichiometric problems.

Review Vocabularychemical reaction: a process in which the atoms of one or more substances are rearranged to form different substances

Stoichiometric Calculations MAIN Idea The solution to every stoichiometric problem requires a balanced chemical equation.

Real-World Reading Link Baking requires accurate measurements. That is why it is necessary to follow a recipe when baking cookies from scratch. If you need to make more cookies than a recipe yields, what must you do?

Using StoichiometryWhat tools are needed to perform stoichiometric calculations? All stoi-chiometric calculations begin with a balanced chemical equation. Mole ratios based on the balanced chemical equation are needed, as well as mass-to-mole conversions.

Stoichiometric mole-to-mole conversion The vigorous reac-tion between potassium and water is shown in Figure 11.3. The balanced chemical equation is as follows.

2K(s) + 2 H 2 O(l) → 2KOH(aq) + H 2 (g)

From the balanced equation, you know that two moles of potassium yields one mole of hydrogen. But how much hydrogen is produced if only 0.0400 mol of potassium is used? To answer this question, identify the given, or known, substance and the substance that you need to determine. The given substance is 0.0400 mol of potassium. The unknown is the number of moles of hydrogen. Because the given sub-stance is in moles and the unknown substance to be determined is also in moles, this problem involves a mole-to-mole conversion.

To solve the problem, you need to know how the unknown moles of hydrogen are related to the known moles of potassium. In Section 11.1, you learned to derive mole ratios from the balanced chemical equation. Mole ratios are used as conversion factors to convert the known number of moles of one substance to the unknown number of moles of another substance in the same reaction. Several mole ratios can be written from the equation, but how do you choose the correct one?

Incorporate information from this section into

your Foldable.

■ Figure 11.3 Potassium metal reacts vigorously with water, releasing so much heat that the hydrogen gas formed in the reaction catches fire.

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Section 11.2 • Stoichiometric Calculations 373


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AssessmentAssessmentKnowledge Have students

consider the reaction of 2.50 mol calcium hydride with excess water according to this equation.Ca H 2 + 2 H 2 O → Ca( OH) 2 + 2H 2 Ask the following questions: What mass of calcium hydroxide will be produced? 185 g What mass of water is needed? 90.1 g What mass of hydrogen is produced? 10.1 g What is the mass of 2.50 mol calcium hydride? 105 g Have students verify the law of conservation of mass. 105 g CaH 2 + 90.1 g H 2 O = 185 g Ca( OH) 2 + 10. 1 g H 2 OL

Differentiated InstructionBelow Level Stress the importance of solving stoichiometric problems step by step and understanding the reason for each step. Ask students why the balanced chemical equation is important. It determines the mole ratios. Give them an equation, and ask them to explain how they would convert from moles of one substance to moles of another substance in the same equation. Use the mole ratio: moles of unknown substance divided by moles of known substance. This might be a point of confusion. Have students write a mole ratio on one side of an index card and the inverse of the same ratio on the other side. As they solve a problem, they can look at the index card to determine which ratio gives the correct unit for the answer. BL

Quick Demo

Varying Reactant Ratios Punch a small hole in the center of the bottom of a potato chip or tennis ball can. Punch another hole in the side of the can an inch from the top. Tape a wood splint to a meter stick. Place the lid on the can and purge the can with hydrogen from a tank of hydrogen using the hole near the lid. Then, fill the can with hydrogen. Place tape over each hole. Place the closed can with its lid on the floor. Quickly remove the tape from both openings and light the top opening with the burning splint. Hydrogen will burn quietly with a blue flame. After several minutes, the hydrogen and oxygen mixture will ignite and the can will launch into the air. Ask students why, at first, the hydrogen burned only at the opening of the can. There was no oxygen in the can, so hydrogen could combine with oxygen only at the opening. Ask why the hydrogen burned quietly, not explosively. There was only a limited amount of hydrogen in a large amount of air. Ask why the can exploded after a time. Oxygen was pulled into the can. OL EL

As shown below, the correct mole ratio, 1 mol H 2 to 2 mol K, has moles of unknown in the numerator and moles of known in the denom-inator. Using this mole ratio converts the moles of potassium to the unknown number of moles of hydrogen.

moles of known × moles of unknown __ moles of known = moles of unknown

0.0400 mol K × 1 mol H 2 _ 2 mol K = 0.0200 mol H 2

The following Example Problems show mole-to-mole, mole-to-mass, and mass-to-mass stoichiometry problems. The process used to solve these problems is outlined in the Problem-Solving Strategy below.

Problem-Solving StrategyMastering StoichiometryThe flowchart below outlines the steps used to solve mole-to-mole, mole-to-mass, and mass-to-mass stoichiometric problems.

Moles of given substance

Mass of given substance Mass of unknown substance

Moles of unknown substance

Step 1

Step 2 Step 4

Step 3

moles of unknownmoles of given

1mol

num

ber o

f gra

ms

num

ber o

f gra

ms

1mol

no direct conversion

Start with a balanced equation.Interpret the equation in terms of moles.

Convert from gramsto moles of the given substance. Use the inverse of themolar mass as theconversion factor.

Convert frommoles of unknownto grams ofunknown. Use themolar mass as theconversion factor.

Convert from moles of the given substance to moles of the unknown substance. Use the appropriate mole ratio from the balanced chemical equation as the conversion factor.

1. Complete Step 1 by writing the balanced chemical equation for the reaction.

2. To determine where to start your calculations, note the unit of the given substance.

• If mass (in grams) of the given substance is the starting unit, begin your calculations with Step 2.

• If amount (in moles) of the given substance is the starting unit, skip Step 2 and begin your calculations with Step 3.

3. The end point of the calculation depends on the desired unit of the unknown substance.

• If the answer must be in moles, stop after completing Step 3.

• If the answer must be in grams, stop after completing Step 4.

Apply the StrategyApply the Problem-Solving Strategy to Example Problems 11.2, 11.3, and 11.4.




IN-CLASS Example

Question Bu tane, C 4 H 10 , is the gas burned in disposable lighters. How many moles of oxygen are needed to burn 5.0 mol of butane in a lighter to produce carbon dioxide and water?

Answer2 C 4 H 10 + 1 3 O 2 → 8 CO 2 + 10 H 2 O

13 mol O 2

__ 2 mo l C 4 H 10

5.0 m ol C 4 H 10 × 13 mol O 2

__ 2 mol C 4 H 10

= 32.5 mol O 2


11. a. 2 CH 4 (g) + S 8 (s) → 2CS 2 (l) + 4H 2 S(g)

b. 1.50 m ol S 8 × 2 mol CS 2

__ 1 mol S 8

= 3.00 mol C S 2

c. 1.50 m ol S 8 × 4 mol H 2 S

__ 1 mol S 8

= 6.00 mol H 2 S 12. a. 2SO 2 (g ) + O 2 (g) + 2H 2 O(l) →

2 H 2 SO 4 (aq)

b. 12.5 m ol SO 2 × 2 mo l H 2 SO 4

__ 2 mol S O 2

= 12.5 m ol H 2 SO 4 produced

c. 12.5 m ol SO 2 ×1 mol O 2 __

2 mol SO 2 = 6. 25 mol O 2 needed

Chemistry JournalHow much acid is neutralized? Have students find the active ingredients on the label of a container of antacid tablets and identify the compound that neutralizes stomach acid. Have them note the recommended dosage and the amount of the neu-tralizing ingredient per tablet. Ask students to determine the mass of the active ingre-dient per dose. If the active ingredient is a metal hydroxide, the hydroxide reacts with HCl to produce a metal chloride and water. If the active ingredient is a metal carbon-ate, the products are a metal chloride, carbon dioxide, and water. Have students write the equation of the reaction of the antacid. Using the mass of the active ingredient per dose, have them determine the mass of hydrochloric acid neutralized per dose. OL

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Mole-to-Mole Stoichiometry One disadvantage of burning propane ( C 3 H 8 ) is that carbon dioxide (C O 2 ) is one of the products. The released carbon dioxide increases the concentration of C O 2 in the atmosphere. How many moles of C O 2 is produced when 10.0 mol of C 3 H 8 is burned in excess oxygen in a gas grill?

1 Analyze the ProblemYou are given moles of the reactant, C3H8 and must find the moles of the product, C O2. First write the balanced chemical equation, then convert from moles of C3H8 to moles of C O2. The correct mole ratio has moles of unknown substance in the numerator and moles of known substance in the denominator.

Knownmoles C3H8 = 10.0 mol C3H8

Unknownmoles CO2 = ? mol CO2

2 Solve for the UnknownWrite the balanced chemical equation for the combustion of C3H8.Use the correct mole ratio to convert moles of known ( C3H8) to moles of unknown (C O2).

10.0 mol ? molC3H8(g) + 5 O2(g) → 3C O2(g) + 4 H2O(g)

Mole ratio:3 mol C O2_1 mol C3H8

10.0 mol C3H8 ×3 mol C O2_1 mol C3H8

= 30.0 mol CO2

Burning 10.0 moles of C3H8 produces 30.0 moles C O2.

3 Evaluate the AnswerBecause the given number of moles has three significant figures, the answer also has three figures. The balanced chemical equation indi-cates that 1 mol of C3H8 produces 3 mol of C O2. Thus, 10.0 mol of C3H8

produces three times as many moles of C O2, or 30.0 mol.

PRACTICE Problems Extra Practice Page 983 and glencoe.com

11. Methane and sulfur react to produce carbon disulfide (C S2), a liquid often used in the production of cellophane.

___C H4(g) + ___ S8(s)→ ___C S2(l) + ___ H2S(g)

a. Balance the equation.

b. Calculate the moles of C S2 produced when 1.50 mol S8 is used.

c. How many moles of H2S is produced?

12. Challenge Sulfuric acid ( H2SO4) is formed when sulfur dioxide (S O2)reacts with oxygen and water.

a. Write the balanced chemical equation for the reaction.

b. How many moles of H2SO4 is produced from 12.5 moles of S O2?

c. How many moles of O2 are needed?

Real-World Chemistry Outdoor Cooking

Gas Grills Using outdoor grills is a popular way to cook. Gas grills burn either natural gas or propane that is mixed with air. The initial spark is provided by a grill starter. Propane is more commonly used for fuel because it can be supplied in liquid form in a portable tank. Combustion of liquid propane also releases more energy than natural gas.

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IN-CLASS Example

Question What is the mass of hydrogen produced when a 0.200 mol sample of sodium reacts with an excess of water to produce hydrogen and sodium hydroxide?

Answer2 Na + 2 H2O → 2 NaOH + H21 m ol H 2

__ 2 mol Na

0.200 mol Na × 1 mol H 2

__ 2 mol Na

= 0.100 mol H2

0.10 0 mol H 2 × 2.016 g H 2

__ 1 mol H 2

= 0.202 g H 2 will be produced


13. 88.6 g Cl 2 14. a. 177 g Cl 2 b. 15.0 g C c. 292 g 15. 64.6 4 g N 2 16. 2SO 2 (g) + O 2 (g) + 2H 2 O(l) →

2 H 2 SO 4 (aq); 3.83 g H 2 SO 4

See Overset

Cultural Diversity

Stoichiometry in Soap Making Soap making dates to 2800 b.c. in Babylonia. During the 17th century, soap was a luxury item for the wealthy. However, as materials became more readily available, soap was made in almost every household. Crude stoi-chiometric techniques were used by all early soap-making cultures. Soap making requires three steps, making lye, rendering (or clean-ing) fat, and then heating the lye-fat mixture. Sodium chloride was added to make the soap

hard. Scents were also added, and the soap was rolled into balls or cut into bars to be stored. Deer fat, whale blubber, tallow, and olive oil have been used as sources of fat in soap depending upon which was available in the area. Countries bordering the Mediterranean Sea used the ashes of local shrubs instead of potash (K 2 CO 3 ). Ashes from these plants are rich in sodium carbon-ate, and were a valuable trade item called barilla.


Mole-to-Mass Stoichiometry Determine the mass of sodium chloride (NaCl), commonly called table salt, produced when 1.25 mol of chlorine gas (C l 2 ) reacts vigorously with excess sodium.

1 Analyze the ProblemYou are given the moles of the reactant, C l2, and must determine the mass of the product, NaCl. You must convert from moles of C l2 to moles of NaCl using the mole ratio from the equation. Then, you need to convert moles of NaCl to grams of NaCl using the molar mass as the conversion factor.

Known Unknownmoles of chlorine = 1.25 mol C l2 mass of sodium chloride = ? g NaCl

2 Solve for the Unknown 1.25 mol ? g2Na(s) + Cl2(g) → 2NaCl(s) Write the balanced chemical equation, and

identify the known and the unknown values.

Mole ratio: 2 mol NaCl_1 mol C l2

1.25 mol C l2 ×2 mol NaCl_1 mol C l2

= 2.50 mol NaCl Multiply moles of C l2 by the mole ratio to get moles of NaCl.

2.50 mol NaCl ×58.44 g NaCl_1 mol NaCl

= 146 g NaCl Multiply moles of NaCl by the molar mass to get grams of NaCl.

3 Evaluate the AnswerBecause the given number of moles has three significant figures, the mass of NaCl also has three. To quickly assess whether the calculated mass value for NaCl is correct, perform the calculations in reverse: divide the mass of NaCl by the molar mass of NaCl, and then divide the result by 2. You will obtain the given number of moles of C l2.


13. Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How much chlorine gas, in grams, is obtained from the process diagrammed at right?

14. Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride ( TiC l4) is extracted from titanium oxide ( Ti O2) using chlorine and coke (carbon).

Ti O2(s) + C(s) + 2C l2(g) → TiC l4(s) + C O2(g)

a. What mass of C l2 gas is needed to react with 1.25 mol of Ti O2?

b. What mass of C is needed to react with 1.25 mol of Ti O2?

c. What is the mass of all of the products formed by reaction with 1.25 mol of Ti O2?

Calculations with Significant Figures

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Math Handbook

Stoichiometric mole-to-mass conversion Now, suppose you know the number of moles of a reactant or product in a reaction and you want to calculate the mass of another product or reactant. This is an example of a mole-to-mass conversion.

2.50 mol ? g

Electricenergy NaCl

Cl2

Na




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IN-CLASS Example

Question The Titan rocket uses a combination of hydrazi ne, N 2 H 4 , and dinitrogen tetrox ide , N 2 O 4 as rocket fuel according to the reaction:2 N 2 H 4 + N 2 O 4 → 3N 2 + 4H 2 OIf 200 g of hydrazine is used, how many grams of nitrogen is produced?

Answer 2 N 2 H 4 + N 2 O 4 → 3 N 2 + 4 H 2 O

200. g N 2 H 4 × 1 mo l N 2 H 4

__ 32.05 g N 2 H 4

= 6.24 mo l N 2 H 4

6.24 mol N 2 H 4 × 4 mol H 2 O

__ 2 mol N 2 H 4

= 12.5 mol H 2 O

12.5 m ol H 2 O × 18.02 g H 2 O

__ 1 mol H 2 O

= 225 g H 2 O

3 AssessCheck for UnderstandingWrite the balanced equation for the reaction between iron and sulfur on the board: 16Fe + 3 S 8 → 8F e 2 S 3 . Have students determine the mass of iron(III) sulfide produced if 7.00 g of iron is used. 13.0 g F e 2 S 3 OL

ReteachWrite this equation on the board: 6C O 2 + 6 H 2 O → C 6 H 12 O 6 + 6 O 2 . Ask students how many grams of glucose form when 88.0 g of C O 2 reacts? Before they pick up their pencils and calculators, have them verbalize each step in the solution and explain why each step is needed. Once they can explain each step, they will be able to solve the prob-lem. 60.0 g C 6 H 12 O 6 OL

ExtensionGive students groups a stoichiomet-ric problem to solve. Ask them to illustrate the solution using a poster similar to the Problem-Solving Strategy graphic on p. 374. OL

Video Lab

DVD Stoichiometry

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Stoichiometric mass-to-mass conversion If you were prepar-ing to carry out a chemical reaction in the laboratory, you would need to know how much of each reactant to use in order to produce the mass of product you required. Example Problem 11.4 demonstrates how you can use a measured mass of the known substance, the balanced chemi-cal equation, and mole ratios from the equation to find the mass of the unknown substance. The ChemLab at the end of this chapter will pro-vide you with laboratory experience in determining a mole ratio.

100.0 g NaN3 → ? g N2(g)

N2 gas


Mass-to-Mass Stoichiometry Ammonium nitrate (N H 4 N O 3 ), an important fertilizer, produces dinitrogen oxide ( N 2 O) gas and H 2 O when it decomposes. Determine the mass of H 2 O produced from the decomposition of 25.0 g of solid N H 4 N O 3 .

1 Analyze the ProblemYou are given a description of the chemical reaction and the mass of the reactant. You need to write the balanced chemical equation and convert the known mass of the reactant to moles of the reactant. Then, use a mole ratio to relate moles of the reactant to moles of the product. Finally, use the molar mass to convert from moles of the product to the mass of the product.

Known Unknownmass of ammonium nitrate = 25.0 g N H4NO3 mass of water = ? g H2O

2 Solve for the Unknown 25.0 g ? gNH4NO3(s)→ N2O(g) + 2 H2O(g) Write the balanced chemical equation, and

identify the known and unknown values.

25.0 g N H4NO3 ×1 mol N H4NO3__

80.04 g N H4NO3= 0.312 mol N H4NO3

Multiply grams of N H4NO3 by the inverse of molar mass to get moles of N H4NO3.

Mole ratio: 2 mol H2O__1 mol N H4NO3

0.312 mol N H4NO3 ×2 mol H2__

1 mol N H4NO3= 0.624 mol H2O

Multiply moles of N H4NO3 by the mole ratio to get moles of H2O.

0.624 mol H2O ×18.02 g H2O_1 mol H2O

= 11.2 g H2OMultiply moles of H2O by the molar mass to get grams of H2O.

3 Evaluate the AnswerThe number of significant figures in the answer, three, is determined by the given moles of NH4NO3. To verify that the mass of H2O is correct, perform the calculations in reverse.


15. One of the reactions used to inflate automobile air bags involves sodium azide (Na N3): 2Na N3(s)→ 2Na(s) + 3 N2(g). Determine the mass of N2 produced from the decomposition of NaN 3 shown at right.

16. Challenge In the formation of acid rain, sulfur dioxide (S O2) reacts with oxygen and water in the air to form sulfuric acid ( H2SO4). Write the balanced chemical equation for the reaction. If 2.50 g of S O2 reacts with excess oxygen and water, how much H2SO4, in grams, is produced?

Dimensional Analysispage 956

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See the MiniLab worksheet in your FAST FILE.


Purpose Predict the mass of a product and compare it with the experimental mass.

Process Skills collect and interpret data, predict, use numbers

Safety Precautions Approve lab safety forms before beginning. Warn students about the hot crucible.

Disposal The sodium carbonate product can be placed in the trash.

Teaching StrategiesHave students calculate the final mass of the product before starting

Expected Results 3.12 g of baking soda yields about 1.90 g Na 2 CO 3

Analysis 1. During heating, the product initially

looked “wet” and bubbles appeared. In time, the product “dried out.”

2. The two masses should be similar. 3. Assuming accepted and actual values are

1.97 g and 1.90 g, respectively; error = −0.07 g; % error = 3.55%.

4. errors from each mass measurement, weight of moisture absorbed by crucible OL

Section 11.211.2 Assessment 17. The coefficients in the balanced equation indicate the molar relationship

between each pair of reactants and products. 18. balance the equation; convert the mass of the known substance to moles of

known substance; use the mole ratio to convert from moles of the known to moles of the unknown; convert moles of unknown to mass of the unknown

19. moles of unknown/moles of known

20. Write a balanced equation. Convert the given mass of Mg to moles. Use the mole ratio from the balanced equation to convert moles of Mg to moles of Br. Convert from moles of Br to mass of Br.

21. 15.2 g NH 3 22. Concept maps will vary, but all should show the use of these conversion

factors: the inverse of molar mass, the mole ratio, the molar mass.

Set Overset


Section 11.211.2 Assessment

Apply StoichiometryHow much sodium carbonate (N a 2 C O 3 ) is pro-duced when baking soda decomposes? Baking soda is used in many baking recipes because it makes batter rise, which results in a light and fluffy texture. This occurs because baking soda, sodium hydrogen carbonate (NaHC O 3 ), decomposes upon heating to form carbon dioxide gas according to the following equation.

2NaHC O 3 → N a 2 C O 3 + C O 2 + H 2 O

Procedure

1. Read and complete the lab safety form.

2. Create a data table to record your experimental data and observation.

3. Use a balance to measure the mass of a clean, dry crucible. Add about 3.0 g of sodium hydro-gen carbonate (NaHC O 3 ), and measure the combined mass of the crucible and NaHC O 3 . Record both masses in your data table, and calculate the mass of the NaHC O 3 .

4. Use this starting mass of NaHC O 3 and the bal-anced chemical equation to calculate the mass of NaHC O 3 that will be produced.

5. Set up a ring stand with a ring and clay triangle for heating the crucible.

6. Heat the crucible with a Bunsen burner, slowly at first and then with a stronger flame, for 7–8 min. Record your observations during the heating.

7. Turn off the burner, and use crucible tongs to remove the hot crucible.WARNING: Do not touch the hot crucible with your hands.

8. Allow the crucible to cool, and then measure the mass of the crucible and NaHC O 3 .

Analysis1. Describe what you observed during the heating

of the baking soda.2. Compare your calculated mass of NaHC O 3

with the actual mass you obtained from the experiment.

3. Calculate Assume that the mass of N a 2 HC O 3 that you calculated in Step 4 is the accepted value for the mass of product that will form. Calculate the error and percent error associated with the experimentally measured mass.

4. Identify sources of error in the procedure that led to errors calculated in Question 3.

Section Summary◗ ◗ Chemists use stoichiometric calcula-

tions to predict the amounts of reactants used and products formed in specific reactions.

◗ ◗ The first step in solving stoichiometric problems is writing the balanced chemical equation.

◗ ◗ Mole ratios derived from the balanced chemical equation are used in stoichiometric calculations.

◗ ◗ Stoichiometric problems make use of mole ratios to convert between mass and moles.

17. MAIN Idea Explain why a balanced chemical equation is needed to solve a stoichiometric problem.

18. List the four steps used in solving stoichiometric problems.

19. Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometric problem.

20. Apply How can you determine the mass of liquid bromine (B r 2 ) needed to react completely with a given mass of magnesium?

21. Calculate Hydrogen reacts with excess nitrogen as follows:

N 2 (g) + 3 H 2 (g) → 2N H 3 (g)

If 2.70 g of H 2 reacts, how many grams of N H 3 is formed?

22. Design a concept map for the following reaction.

CaC O 3 (s) + 2HCl(aq) → CaC l 2 (aq) + H 2 O(l) + C O 2 (g)

The concept map should explain how to determine the mass of CaC l 2 produced from a given mass of HCl.

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MAIN IdeaLimited Reactants Refer students to the Launch Lab and ask what hap-pened to the potassium permanga-nate. The solution turned from purple to colorless. How much sodium hydrogen sulfite was added before the color change occurred? Answers will vary. It should be around 8 mL of N aHSO 3 added to change the color. Ask students what happened if they added another drop of sodium hydrogen sulfite. Nothing, the solution remained colorless. Ask if they can explain why that was the case. All the potassium permanganate had reacted, and there was no more to react with the addi-tional drop. OL

2 Teach

■ Caption Question Fig. 11.4 one additional hammer is needed BL EL

Quick Demo

Adding Reactants Using a straw, exhale into a test tube of limewater. Explain that the lime-water becomes cloudy because insoluble calcium carbonate is formed. Write the equation for the reaction on the board or project the image: C O 2 (g) + CaO(aq) → CaC O 3 (s). Ask stu-dents to determine the mass of calcium carbonate that would be produced if you exhaled 0.0900 mol CO 2 into the test tube. 9.00 g Ca CO 3 OL

Chemistry JournalStoichiometric Steps In their chemistry journals, have students write the heading “Steps in Solving Limiting Reactant Stoichiometric Problems.” As they study section 11.3, have students list the steps needed for calculating the amounts of products formed when the masses of all reactants are given. Students should explain the purpose for each step. OL

Section 11.311.3

Objectives

◗◗ Identify the limiting reactant in a chemical equation.

◗◗ Identify the excess reactant, and calculate the amount remaining after the reaction is complete.

◗◗ Calculate the mass of a product when the amounts of more than one reactant are given.

Review Vocabularymolar mass: the mass in grams of one mole of any pure substance

New Vocabularylimiting reactantexcess reactant

Limiting ReactantsMAIN Idea A chemical reaction stops when one of the reactants is used up.

Real-World Reading Link If there are more boys than girls at a school dance, some boys will be left without dance partners. The situation is much the same for the reactants in a chemical reaction—excess reactants cannot participate.

Why do reactions stop?Rarely in nature are the reactants present in the exact ratios specified by the balanced chemical equation. Generally, one or more reactants are in excess and the reaction proceeds until all of one reactant is used up. When a reaction is carried out in the laboratory, the same principle applies. Usually, one or more of the reactants are in excess, while one is limited. The amount of product depends on the reactant that is limited.

Limiting and excess reactants Recall the reaction from the Launch Lab. After the colorless solution formed, adding more sodium hydrogen sulfite had no effect because there was no more potassium permanganate available to react with it. Potassium permanganate was a limiting reactant. As the name implies, the limiting reactant limits the extent of the reaction and, thereby, determines the amount of product formed. A portion of all the other reactants remains after the reaction stops. Reactants leftover when a reaction stops are excess reactants.

To help you understand limiting and excess reactants, consider the analogy in Figure 11.4. From the available tools, four complete sets consisting of a pair of pliers, a hammer, and two screwdrivers can be assembled. The number of sets is limited by the number of available hammers. Pliers and screwdrivers remain in excess.

Available tools

Sets of tools

Extra tools

Set 1 Set 2 Set 3 Set 4

■ Figure 11.4 Each tool set must have one hammer, so only four sets can be assembled. Interpret How many more hammers are required to complete a fifth set?

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Section 11.3 • Limiting Reactants 379


Reading Check six molecules

Chemistry ProjectMaximum Yield of Carbon Dioxide Have student groups design an experiment to determine the maximum yield of carbon dioxide produced by the reaction of bak-ing soda with vinegar. Suggested equipment includes a plastic soft drink bottle, vine-gar, baking soda, a balloon to cover the mouth of the bottle, and a teaspoon. One teaspoonful of baking soda is approximately 10 g and one teaspoonful of vinegar is approximately 4.9 mL. Vinegar is 5% acetic acid. Suggest that students use a constant amount of baking soda (one teaspoonful) and vary the amount of vinegar added. Ask students to find out if there is a point at which no more product is produced. OL COOP LEARN

IdentifyMisconceptions

Students often have difficulty determining the limiting reactant when given the mass of more than one reactant.Uncover the Misconception Ask how much magnesium oxide is produced when 48.6 g of mag-nesium reacts with 64.0 g of oxygen according to this equation: 2Mg + O 2 → 2MgO. Have students determine the moles of each of the reactants given in the problem. 2.00 mol Mg, 2.00 m ol O 2 Ask which is the limiting reactant. Mg Ask what is the mass of magnesium oxide produced. 80.6 g Demonstrate the Concept Use nuts and bolts to represent moles of atoms. Represent 2 mol of Mg using two bolts and 2 mol of O 2 using four nuts. Have stu-dents identify the mole ratio betwee n O 2 molecules and Mg atoms. 1 mol O 2 molecules/2 mol Mg atoms, or 2 nuts/2 bolts Ask what is in excess. 1 m ol O 2 , or 2 nuts Represent 1 mol of MgO using one bolt and one nut. Ask how many MgO for-mula units can be formed. 2 units of 1 nut and 1 bolt with 2 nuts left overAssess New Knowledge Give students this problem. A 54.0-g sample of aluminum reacts with 508 g of iodine according to this equation: 2Al + 3 I 2 → 2A lI 3 . Have them determine the number of moles of each reactant. 2.00 mol Al, 2.00 mo l I 2 Ask which is the limiting reactant. I 2 Ask how many moles of Al I 3 are formed. 1.33 mol OL BL EL

Interactive Figure Students can interact with the animation at glencoe.com.

Determining the limiting reactant The calculations you did in the previous section were based on having the reactants present in the ratio described by the balanced chemical equation. When this is not the case, the first thing you must do is determine which reactant is limiting.

Consider the reaction shown in Figure 11.5, in which three mole-cules of nitrogen ( N 2 ) and three molecules of hydrogen ( H 2 ) react to form ammonia (N H 3 ). In the first step of the reaction, all the nitrogen molecules and hydrogen molecules are separated into individual atoms. These atoms are available for reassembling into ammonia molecules, just as the tools in Figure 11.4 are available to be assembled into tool kits. How many molecules of ammonia can be produced from the avail-able atoms? Two ammonia molecules can be assembled from the hydro-gen atoms and nitrogen atoms because only six hydrogen atoms are available—three for each ammonia molecule. When the hydrogen is gone, two unreacted molecules of nitrogen remain. Thus, hydrogen is the limiting reactant and nitrogen is the excess reactant. It is important to know which reactant is the limiting reactant because, as you have just read, the amount of product formed depends on this reactant.

Reading Check Extend How many more hydrogen molecules would be needed to completely react with the excess nitrogen molecules shown in Figure 11.5?

Calculating the Product when a Reactant Is LimitingHow can you calculate the amount of product formed when one of the reactants is limiting? Consider the formation of disulfur dichloride ( S 2 C l 2 ), which is used to vulcanize rubber. As shown in Figure 11.6, the properties of vulcanized rubber make it useful for many products. In the production of disulfur dichloride, molten sulfur reacts with chlorine gas according to the following equation.

S 8 (l) + 4C l 2 (g) → 4 S 2 C l 2 (l)

If 200.0 g of sulfur reacts with 100.0 g of chlorine, what mass of disulfur dichloride is produced?

Calculating the limiting reactant The masses of both reactants are given. First, determine which one is the limiting reactant, because the reaction stops producing product when the limiting reactant is used up.

Three nitrogen molecules(six nitrogen atoms)

Before Reaction After Reaction

Three hydrogen molecules(six hydrogen atoms)

Two ammonia molecules(two nitrogen atoms, six hydrogen atoms)

Two nitrogen molecules(four nitrogen atoms)

++

■ Figure 11.5 If you check all the atoms present before and after the reaction, you will find that some of the nitrogen molecules are unchanged. These nitrogen molecules are the excess reactant.

■ Figure 11.6 Natural rubber, which is soft and very sticky, is hardened in a chemical process called vulcanization. During vulcanization, molecules become linked together, forming a durable material that is harder, smoother, and less sticky. These properties make vulcanized rubber ideal for many products, such as this caster.

Interactive Figure To see an animation of limiting reactants, visit glencoe.com.




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AssessmentAssessmentKnowledge Give students

this recipe for chocolate chip cookies.2 1 _ 4 c flour1 tsp salt1 tsp baking soda1 c softened butter (2 sticks)3 _ 4 c granulated sugar3 _ 4 c brown sugar

2 eggs2 tsp vanilla3 oz chocolate chipsYields 4 dozen cookiesAsk students how they could make twice as many cookies. double the ingredients Ask how they could make these cookies with only one egg. cut the recipe in half Ask students to relate this to limiting reactants. The number of eggs limits the number of cookies that can be made. Likewise, the amounts of reactants determines the quantity of products. OL

Content BackgroundTesting for Chemicals Prior to 1940, the primary pigment in lead-based paints was PbC O 3 ·P b(OH) 2 . Today, we know that lead in household products poses a health hazard, but old paint is still found in many homes. The level of lead in homes can be determined using a lead detection kit. One lead test kit uses a swab containing potassium rhodizonate ( C 6 K 2 O 6 ). The swab is rubbed over the surface to be tested. If lead is present, a pink color appears on the swab due to the presence of lead(II) rhodizonate. The reaction is a single replacement described by the following equatio n: Pb 2+ + C 6 K 2 O 6 → C 6 PbO 6 + 2 K +The larger the amount of lead, the more red the swab becomes. This kit can detect amounts of lead as low as 0.006 g.


Advanced Learners Divide students into groups and tell them they are to form their own cookie company. They have the following ingredients available for making the cookies described in the Assessment on this page. Note, 1 teaspoon = 4.9 mL.

1000 c flour 10 doz eggs 40 sticks butter1.00 L baking soda 100 oz chocolate chips 500 c brown sugar500 c granulated sugar 1.00 L salt 1.00 L vanilla

Which of the ingredients is the limiting ingredient? butter How many dozen cookies can be made? 80 dozen If a dozen cookies sells for $4.00, how much money would the company gross? $320.00 AL COOP LEARN

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Moles of reactants Identifying the limiting reactant involves find-ing the number of moles of each reactant. You can do this by converting the masses of chlorine and sulfur to moles. Multiply each mass by a con-version factor that relates moles and mass—the inverse of molar mass.

100.0 g C l 2 × 1 mol C l 2 _ 70.91 g C l 2 = 1.410 mol C l 2

200.0 g S 8 × 1 mol S 8 _ 256.5 g S 8 = 0.7797 mol S 8

Using mole ratios The next step involves determining whether the two reactants are in the correct mole ratio, as given in the balanced chemical equation. The coefficients in the balanced chemical equation indicate that 4 mol of chlorine is needed to react with 1 mol of sulfur. This 4:1 ratio from the equation must be compared with the actual ratio of the moles of available reactants just calculated above. To determine the actual ratio of moles, divide the number of available moles of chlo-rine by the number of available moles of sulfur.

1.410 mol C l 2 available __ 0.7797 mol S 8 available = 1.808 mol C l 2 available __ 1 mol S 8 available

Only 1.808 mol of chlorine is available for every 1 mol of sulfur, instead of the 4 mol of chlorine required by the balanced chemical equation. Therefore, chlorine is the limiting reactant.

Calculating the amount of product formed After determin-ing the limiting reactant, the amount of product in moles can be calcu-lated by multiplying the given number of moles of the limiting reactant (1.410 mol C l 2 ) by the mole ratio relating disulfur dichloride and chlo-rine. Then, moles of S 2 C l 2 is converted to grams of S 2 C l 2 by multiplying by the molar mass. These calculations can be combined as shown.

1.410 mol C l 2 × 4 mol S 2 C l 2 _ 4 mol C l 2 × 135.0 g S 2 C l 2

_ 1 mol S 2 C l 2 = 190.4 g S 2 C l 2

Thus, 190.4 g S 2 C l 2 forms when 1.410 mol C l 2 reacts with excess S 8 .

Analyzing the excess reactant Now that you have determined the limiting reactant and the amount of product formed, what about the excess reactant, sulfur? How much of it reacted?

Moles reacted You need to make a mole-to-mass calculation to determine the mass of sulfur needed to react completely with 1.410 mol of chlorine. First, obtain the number of moles of sulfur by multiplying the moles of chlorine by the S 8 -to-C l 2 mole ratio.

1.410 mol C l 2 × 1 mol S 8 _ 4 mol C l 2 = 0.3525 mol S 8

Mass reacted Next, to obtain the mass of sulfur needed, multiply 0.3525 mol of S 8 by its molar mass.

0.3525 mol S 8 × 265.5 g S 8

_ 1 mol S 8 = 90.42 g S 8 needed

Excess remaining Knowing that 200.0 g of sulfur is available and that only 90.42 g of sulfur is needed, you can calculate the amount of sulfur left unreacted when the reaction ends.

200.0 g S 8 available - 90.42 g S 8 needed = 109.6 g S 8 in excess

Careers In chemistry

Pharmacist Knowledge of drug composition, modes of action, and possible harmful interactions with other substances allows a pharma-cist to counsel patients on their care. Pharmacists also mix chemicals to form powders, tablets, ointments, and solutions. For more information on chemistry careers, visit glencoe.com.

VOCABULARYSCIENCE USAGE V. COMMON USAGE

ProductScience usage: a new substance formed during a chemical reactionThe sole reaction product was a colorless gas.

Common usage: something producedThe cosmetics counter in the depart-ment store had hundreds of products from which to choose.

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IN-CLASS Example

Question When silver (Ag) reacts with sulfur (S 8 ,), the compound sil-ver sulfide (Ag 2 S) is formed.16Ag + S 8 → 8Ag 2 S a. When 4.00 g of silver reacts with

4.00 g of sulfur, what mass of sil-ver sulfide is produced?

b. How much of the excess reactant remains after the reaction stops?

Answer a. 4.00 g Ag × 1 mol __

107.9 g Ag

= 0.0371 mol Ag

4.00 g S 8 × 1 mol __ 256.5 g S 8

= 0.0156 mol S 8

0.0371 mol Ag × 8 mol Ag 2 S

__ 16 mol Ag

= 0.0186 mol A g 2 S

0.0186 mo l Ag 2 S × 247.9 g Ag 2 S

__ 1 mol Ag 2 S

= 4.60 g A g 2 S produced.

b. 0.0371 mol Ag × 1 mol S 8

__ 16 mol Ag

= 0.00232 mol S 8 needed

0.00232 mo l S 8 × 256.5 g S 8

__ 1 mol S 8

= 0.595 g S 8 4.00 g - 0.595 g = 3.40 g S 8 in excess

See Overset

DemonstrationLimiting ReactantsPurposeto observe the effect of a limiting reactant on a chemical reactionMaterialsBurets (2); metric ruler; 25 mL graduated cylinder; stirring rods; 18 × 150 mm test tubes (8); test-tube rack; 0.10M CuS O 4 ; 0.20M KOH; centrifuge; distilled water;

See page 48T for preparation of solutions.Safety Precautions

DisposalCu SO 4 and KOH solutions can be evaporated to dryness in a fume hood. The solid can be disposed of at an approved waste facility.ProcedureFill one buret with 0.20M KOH and the other with 0.10M CuS O 4 . Label eight test tubes

according to the amount of KOH to be added: 2.0 mL, 4.0 mL, 6.0 mL, 8.0 mL, 10.0 mL, 12.0 mL, 14.0 mL, and 16.0 mL. Dispense 10.0 mL C uSO 4 into all eight labeled test tubes. Add KOH according to the volume on the test tube label. Add enough water to each test tube so that each will have 26.0 mL. Stir and place each test tube in the centrifuge for about one min-ute. Remove each test tube from the centrifuge. Measure the height of the precipitate and note the color of the supernatant in each test tube.


Determining the Limiting Reactant The reaction between solid white phosphorus ( P 4 ) and oxygen produces solid tetraphosphorus decoxide ( P 4 O 10 ). This compound is often called diphosphorus pentoxide because its empirical formula is P 2 O 5 .a. Determine the mass of P 4 O 10 formed if 25.0 g of P 4 and 50.0 g of oxygen

are combined.b. How much of the excess reactant remains after the reaction stops?

1 Analyze the ProblemYou are given the masses of both reactants, so you must identify the limiting reactant and use it to find the mass of the product. From moles of the limiting reactant, the moles of the excess reactant used in the reaction can be determined. The number of moles of the excess reactant that reacted can be converted to mass and subtracted from the given mass to find the amount in excess.

Known Unknownmass of phosphorus = 25.0 g P4 mass of tetraphosphorus decoxide = ? g P4O10

mass of oxygen = 50.0 g O2 mass of excess reactant = ? g excess reactant

2 Solve for the UnknownDetermine the limiting reactant. 25.0 g 50.0 g ? g Write the balanced chemical equation, and

identify the known and the unknown.P4(s) + 5 O2(g) → P4O10(s)

Determine the number of moles of the reactants by multiplying each mass by the conversion factor that relates moles and mass—the inverse of molar mass.

25.0 g P4 ×1 mol P4_

123.9 g P4= 0.202 mol P4 Calculate the moles of P4.

50.0 g O2 ×1 mol O2_

32.00 g O2= 1.56 mol O2 Calculate the moles of O2.

Calculate the actual ratio of available moles of O2 and available moles of P4.

1.56 mol O2_0.202 mol P4

=7.72 mol O2_

1 mol P4

Calculate the ratio of moles of O2 to moles of P4.

Determine the mole ratio of the two reactants from the balanced chemical equation.

Mole ratio:5 mol O2_ mol P4

Because 7.72 mol of O2 is available but only 5 mol is needed to react with 1 mol of P4, O2

is in excess and P4 is the limiting reactant. Use the moles of P4 to determine the moles of P4O10 that will be produced. Multiply the number of moles of P4 by the mole ratio of P4O10

(the unknown) to P4 (the known).

0.202 mol P4 ×1 mol P4O10_

1 mol P4= 0.202 mol P4O10 Calculate the moles of product (P4O10) formed.

To calculate the mass of P4O10, multiply moles of P4O10 by the conversion factor that relates mass and moles—molar mass.

0.202 mol P4O10 ×283.9 g P4O10__1 mol P4O10

= 57.3 g P4O10 Calculate the mass of the product P4O10.

Dimensional Analysispage 956

Math Handbook




AssessmentAssessmentKnowledge Have students

write the equation for the reaction of lithium with bromine to produce lithium bromide. Have them deter-mine the following, given that 25.0 g of lithium and 25.0 g of bromine are present at the beginning of the reac-tion: the limiting reactant, the mass of lithium bromide produced, the excess reactant, the mass of the excess reactant. Bromine is the limiting reactant; 27.1 g LiBr; lithium is the excess reactant; 22.8 g excess. OL


23. a. F e 2 O 3 b. Na c. 69.92 g Fe d. 13.6 g Na 24. a. 6C O 2 (g) + 6H 2 O(l) →

C 6 H 12 O 6 (aq) + 6 O 2 (g) b. C O 2 c. H 2 O d. 28.0 g e. 60.0 g

AssessmentAssessmentKnowledge Explain why

the amount of precipitate levels off in the test tubes that have 10.0 mL or more of KOH. KOH is in excess and C uSO 4 is the limiting reactant. OL

ResultsTest tubes labeled 2, 4, 6, 8 have blue superna-tant and increasing amounts of precipitate. Test tubes labeled 10, 12, 14, 16 have the same amount of precipitate and clear colorless supernatant.

Analysis 1. What is the balanced equation for the

reaction between KOH and CuS O 4 ? 2KOH(aq) + CuS O 4 (aq) → Cu(OH ) 2 (s) + K 2 S O 4 (aq).

2. Explain the blue color of the supernatant in some test tubes. CuS O 4 is in excess in these tubes.

3. In which test tube(s) did the maximum amount of precipitate occur? in those with 10, 12, 14, and 16 mL of potassium hydroxide

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Connection Biology Your body needs vitamins, minerals, and ele-ments in small amounts to facilitate normal metabolic reactions. A lack of these substances can lead to abnormalities in growth, development, and the functioning of your body’s cells. Phosphorus, for example, is an essential element in living systems; phosphate groups occur regularly in strands of DNA. Potassium is needed for proper nerve function, muscle control, and blood pressure. A diet low in potassium and high in sodi-um might be a factor in high blood pressure. Another example is vita-min B-12. Without adequate vitamin B-12, the body is unable to synthesize DNA properly, affecting the production of red blood cells.

Because O2 is in excess, only part of the available O2 is consumed. Use the limiting reactant, P4, to determine the moles and mass of O2 used.

0.202 mol P4 ×5 mol O2_1 mol P4

= 1.01 mol O2Multiply the moles of limiting reactant by the mole ratio to determine moles of excess reactant needed.

Convert moles of O2 consumed to mass of O2 consumed.

1.01 mol O 2 ×32.00 g O2_1 mol O2

= 32.3 g O2 Multiply the moles of O2 by the molar mass.

Calculate the amount of excess O2.50.0 g O2 available - 32.3 g O2 consumed = 17.7 g O2 in excess Subtract the mass of O2 used

from the mass available.

3 Evaluate the AnswerAll values have a minimum of three significant figures, so the mass of P4O10 is correctly stated with three digits. The mass of excess O2 (17.7 g) is found by subtracting two numbers that are accurate to the first decimal place. Therefore, the mass of excess O2

correctly shows one decimal place. The sum of the O2 that was consumed (32.3 g) and the given mass of P4 (25.0 g) is 57.3 g, the calculated mass of the product P4O10.


23. The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag: 6Na(s) + F e2O3(s)→ 3N a2O(s) + 2Fe(s). If 100.0 g of Na and 100.0 g of F e2O3 are used in this reaction, determine the following.

a. limiting reactant

b. reactant in excess

c. mass of solid iron produced

d. mass of excess reactant that remains after the reaction is complete

24. Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose ( C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis.

a. Write the balanced chemical equation for the reaction.

b. Determine the limiting reactant.

c. Determine the excess reactant.

d. Determine the mass in excess.

e. Determine the mass of glucose produced.

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Section 11.311.3 Assessment 25. one of the reactants is used up 26. a. The wood limits the reaction. Oxygen is in excess. The fire will burn only

while wood is present. b. Silver is the limiting reactant. Sulfur is in excess. When a layer of tarnish

covers the silver surface, it prevents the sulfur in the air from reacting. c. A decomposition reaction usually has only one reactant. The reaction is

limited by the amount of baking powder present.

27. a. correct b. Phosphorus is the limiting reactant. c. correct

3 AssessCheck for UnderstandingWrite the equation for coal gasifica-tion, a process that converts coal into methane gas, on the board or project the image. 2C + 2 H 2 O → C H 4 + C O 2 Have students determine the yield of methane if 2000 g of carbon and 2000 g of water are used. 900 g OL

ReteachAsk students to determine the limiting reactant and the excess reactant in these situations. 1. charcoal briquettes burning in

an outdoor grill Charcoal is limiting; oxygen is in excess.

2. a candle burning in a closed container Oxygen is limiting; candle wax is in excess.

Ask students what data should be collected and what calculations must be done to confirm their choices. The number of moles of each reactant, the actual mole ratio, and the ratio according to the balanced equation. OL

ExtensionStudents can design an experiment that could be used to determine the maximum rise of a cupcake using different amounts of baking powder. AL

AssessmentAssessmentSkill Ask each student to

write a balanced equation with at least two reactants and provide masses for each. Working in pairs, students should solve their two problems for the limiting reactant, the excess reactant, and the amount of each product. OL COOP LEARN

Section 11.311.3 AssessmentSection Summary◗ ◗ The limiting reactant is the reactant

that is completely consumed during a chemical reaction. Reactants that remain after the reaction stops are called excess reactants.

◗ ◗ To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation.

◗ ◗ Stoichiometric calculations must be based on the limiting reactant.

25. MAIN Idea Describe the reason why a reaction between two substances comes to an end.

26. Identify the limiting and the excess reactant in each reaction. a. Wood burns in a campfire. b. Airborne sulfur reacts with the silver plating on a teapot to produce tarnish

(silver sulfide). c. Baking powder in batter decomposes to produce carbon dioxide.

27. Analyze Tetraphosphorus trisulphide ( P 4 S 3 ) is used in the match heads of some matches. It is produced in the reaction 8 P 4 + 3 S 8 → 8 P 4 S 3 . Determine which of the following statements are incorrect, and rewrite the incorrect statements to make them correct.

a. 4 mol P 4 reacts with 1.5 mol S 8 to form 4 mol P 4 S 3 . b. Sulfur is the limiting reactant when 4 mol P 4 and 4 mol S 8 react. c. 6 mol P 4 reacts with 6 mol S 8 , forming 1320 g P 4 S 3 .

Why use an excess of a reactant? Many reactions stop while portions of the reactants are still present in the reaction mixture. Because this is inefficient and wasteful, chemists have found that by using an excess of one reactant—often the least expensive one—reac-tions can be driven to continue until all of the limiting reactant is used up. Using an excess of one reactant can also speed up a reaction.

Figure 11.7 shows an example of how controlling the amount of a reactant can increase efficiency. Your lab likely uses the type of Bunsen burner shown in the figure. If so, you know that this type of burner has a control that lets you adjust the amount of air that mixes with the methane gas. How efficiently the burner operates depends on the ratio of oxygen to methane gas in the fuel mixture. When the air is limited, the resulting flame is yellow because of glowing bits of unburned fuel. This unburned fuel leaves soot (carbon) deposits on glassware. Fuel is wasted because the amount of energy released is less than the amount that could have been produced if enough oxygen were available. When sufficient oxygen is present in the combustion mixture, the burner pro-duces a hot, intense blue flame. No soot is deposited because the fuel is completely converted to carbon dioxide and water vapor.

■ Figure 11.7 With insufficient oxygen, the burner on the left burns with a yellow, sooty flame. The burner on the right burns hot and clean because an excess of oxygen is available to react completely with the methane gas.

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MAIN IdeaActual and Theoretical Yield Pop a bag of popcorn and ask stu-dents if all of the popcorn popped. Some of the corn kernels will remain unpopped. Tell students that not all reactants produce products in a chemical reaction. The number of popped kernels in the bag is the actual yield. They can use stoichiometry to determine the theoretical yield. BL OL

2 TeachMath in ChemistryCalculate Percentages Students might have learned that to obtain a percent, they should divide the part by the whole and multiply by 100. Help them to see that when they divide the part by the whole, they are finding out the fraction of the whole that the part is, or part per one. When they multiply by 100, they are finding the parts per 100, or percent. Ask what they would do to find parts per million. Multiply part per one by 1 0 6 . BL

Chemistry ProjectEvaluate Choices Have students develop six multiple-choice questions with four answer choices, one being the correct answer. These questions, with the correct answers, can be used in a class game. OL

Chemistry JournalBatting Averages If students do not know what is meant by a baseball batting average, have them research this subject and write in their journals how a batting average is similar to percent yield. BL EL

Section 11.411.4

Objectives

◗◗ Calculate the theoretical yield of a chemical reaction from data.

◗◗ Determine the percent yield for a chemical reaction.

Review Vocabularyprocess: a series of actions or operations

New Vocabularytheoretical yieldactual yieldpercent yield

Percent YieldMAIN Idea Percent yield is a measure of the efficiency of a chemical reaction.

Real-World Reading Link Imagine that you are practicing free throws and you take 100 practice shots. Theoretically, you could make all 100 shots. In actuality, however, you know you will not make all of the shots. Chemical reactions also have theoretical and actual outcomes.

How much product?While solving stoichiometric problems in this chapter, you might have concluded that chemical reactions always proceed in the laboratory according to the balanced equation and produce the calculated amount of product. This, however, is not the case. Just as you are unlikely to make 100 out of 100 free throws during basketball practice, most reac-tions never succeed in producing the predicted amount of product. Reactions do not go to completion or yield as expected for a variety of reasons. Liquid reactants and products might adhere to the surfaces of their containers or evaporate. In some instances, products other than the intended ones might be formed by competing reactions, thus reduc-ing the yield of the desired product. Or, as shown in Figure 11.8, some amount of any solid product is usually left behind on filter paper or lost in the purification process. Because of these problems, chemists need to know how to gauge the yield of a chemical reaction.

Theoretical and Actual Yields In many of the stoichiometric cal-culations you have performed, you have calculated the amount of prod-uct produced from a given amount of reactant. The answer you obtained is the theoretical yield of the reaction. The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.

A chemical reaction rarely produces the theoretical yield of product. A chemist determines the actual yield of a reaction through a careful experiment in which the mass of the product is measured. The actual yield is the amount of product produced when the chemical reaction is carried out in an experiment.

■ Figure 11.8 Silver chromate is formed when potassium chromate is added to silver nitrate. Note that some of the precipitate is left behind on filter paper. Still more of the precipitate is lost because it adheres to the sides of the beaker.

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Section 11.4 • Percent Yield 385


IN-CLASS Example

Question Zinc can be removed from bronze by placing bronze in hydrochloric acid. The zinc reacts with the hydrochloric acid produc-ing zinc chloride and hydrogen gas, and leaving the copper behind. a. If 25.0 g of zinc are in a sample

of bronze, determine the theo-retical yield of hydrogen gas.

Zn + 2 HCl → ZnC l 2 + H 2 b. If the reaction yields 0.680 g H 2 ,

determine the percent yield.

Answer a. 25.0 g Zn × 1 mol Zn __

65.39 g Zn

= 0.382 mol Zn

1 m ol H 2

__ 1 mol Zn

0.382 mol Zn × 1 mol H 2

__ 1 mol Zn

= 0.382 mo l H 2

0.382 mo l H 2 × 2.016 g H 2

__ 1 m ol H 2

= 0.771 g of H 2 is the theoretical yield

b. 0.680 g H 2 (actual yield)

____ 0.771 g H 2 (theoretical yield)

× 100

= 88.2% yield of H 2


Advanced Learners Have capable students design a spreadsheet that can be used to determine the theoretical yield of a product from a given mass of reactant and the per-cent yield of the reaction when given the actual yield. First, they must determine the theoretical yield of the reaction when given a mass of one of the reactants. Second, they must be able to enter an actual yield and calculate the percent yield of the reac-tions. The spreadsheet can be used as a tutorial for other students. AL

Quick Demo

Calculating Yields Add about 5.0 mL of cooking oil to a flask. Place 20 kernels of pop-corn in the flask and cover with a piece of aluminum foil with holes punched in the top. Place the flask on a hot plate and pop the popcorn. Pour out the pop-corn and count the number of kernels that popped as the actual yield. The theoretical yield is 20 kernels. Ask students to determine the percent yield using the results. Answers will vary depending on the data collected. BL EL

Data Analysis lab

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Percent Yield Solid silver chromate (A g 2 Cr O 4 ) forms when potassium chromate ( K 2 Cr O 4 ) is added to a solution containing 0.500 g of silver nitrate (AgN O 3 ). Determine the theoretical yield of A g 2 Cr O 4 . Calculate the percent yield if the reaction yields 0.455 g of A g 2 Cr O 4 .

1 Analyze the ProblemYou know the mass of a reactant and the actual yield of the product. Write the balanced chemical equation, and calculate theoretical yield by converting grams of AgN O3 to moles of AgN O3, moles of AgN O3 to moles of A g2CrO4, and moles of A g2CrO4 to grams of A g2CrO4. Calculate the percent yield from the actual yield and the theoretical yield.

Known Unknown

mass of silver nitrate = 0.500 g AgN O3 theoretical yield = ? g Ag2CrO4

actual yield = 0.455 g A g2CrO4 percent yield = ? % Ag2CrO4

2 Solve for the Unknown

0.500 g ? g2AgNO3(aq) + K2CrO4(aq)→ A g2CrO4(s) + 2KN O3(aq)

Write the balanced chemical equation, and identify the known and the unknown.

0.500 g AgN O3 ×1 mol AgN O3__

169.9 g AgN O3= 2.94 × 10-3 mol AgN O3 Use molar mass to convert grams

of AgN O3 to moles of AgN O3.

2.94 × 10-3 mol AgN O3 ×1 mol A g2CrO4__2 mol AgN O3

= 1.47 × 10-3 mol A g2CrO4Use the mole ratio to convert moles of AgNO3 to moles of A g2CrO4.

1.47 × 10-3 mol A g2CrO4 ×331.7 g A g2CrO4__

1 mol A g2CrO4= 0.488 g A g2CrO4 Calculate the theoretical yield.

0.455 g A g2CrO4__0.488 g A g2CrO4

× 100 = 93.2% A g2CrO4 Calculate the percent yield.

3 Evaluate the AnswerThe quantity with the fewest significant figures has three, so the percent is correctly stated with three digits. The molar mass of A g2CrO4 is about twice the molar mass of AgN O3, and the ratio of moles of AgN O3 to moles of A g2CrO4 in the equation is 2:1. Therefore, 0.500 g of AgN O3 should produce about the same mass of A g2CrO4. The actual yield of A g2CrO4

is close to 0.500 g, so a percent yield of 93.2% is reasonable.

Percent yield Chemists need to know how efficient a reaction is in producing the desired product. One way of measuring efficiency is by means of percent yield. Percent yield of product is the ratio of the actual yield to the theoretical yield expressed as a percent.

Percent Yield

percent yield = actual yield

__ theoretical yield × 100

The actual yield divided by the theoretical yield multiplied by 100 is the percent yield.

Percentspage 965

Math Handbook

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28. 23.9 g of AlCl3 is the theoretical yield.

29. a. 610.3 g ZnI 2 b. 84.48% yield of ZnI 2 30. a. Cu(s) + 2AgN O 3 (aq) →

2Ag(s) + Cu( NO 3 ) 2 (aq) b. 68.0 g of Ag c. 88.2% yield

Data Analysis LabData Analysis Lab

About the Lab• This lab asks students to determine

the amount of available oxygen in the lunar soil, the theoretical yield of oxygen, and the percent yield using an achievable actual yield.

• Soil on the moon is composed of many different oxides that are rich in oxygen. Scientists are studying if the oxygen in moon rocks can pro-vide astronauts with oxygen.

• NASA has simulated the composi-tion of lunar soil. Scientists are try-ing to achieve a cost-effective way of extracting the oxygen from this soil.

• Refer to Gregory Mone, “Mining the Moon,” Popular Science, July 2006, 38–39, for details.

Think Critically 1. Ti O 2 : 16 g; A l 2 O 3 : 178 g ; SiO 2 : 473 g;

FeO: 105 g; MgO: 96 g; CaO: 114 g; N a 2 O: 7 g; K 2 O: 6 g; MnO: 1.00 g; Cr 2 O 3 : 2.00 g

2. Ti O 2 : 0.00641 kg O 2 ; Al 2 O 3 : 0.0838 kg O 2 ; S iO 2 : 0.252 kg O 2 ; FeO: 0.0234 kg O 2 ; MgO: 0.0381 kg O 2 ; CaO: 0.0325 kg O 2 ; N a 2 O: 0.00181 kg O 2 ; K 2 O: 0.000988 kg O 2 ; MnO: 0.000225 kg O 2 ; C r 2 O 3 : 0.000632 kg O 2

3. Si O 2 yields the most; Mn O the least 4. 0.439 kg of O 2 /1.00 kg lunar soil 5. 0.15 kg/0.439 × 100 = 34%

Cultural DiversityNorbert Rillieux was born in New Orleans in 1806. His father was French a planta-tion owner and his mother was an African-American slave. When Norbert was born, his father had the choice of declaring him free or, as was usually the custom, a slave. Fortunately for Norbert, he was declared free. Even though he was a free man, Norbert experienced racial discrimination throughout his life. Taking advantage of the excellent educational opportunities available to him, Norbert studied science and math and became a chemical engineer. Having grown up observing the dangerous and labor-intensive process of refining sugar, he developed a vastly improved process. His invention, the multiple-effect pan evaporator, revolution-ized sugar refining and was eventually patented. His new method could produce a higher-quality sugar at nearly half the cost.

Data Analysis labData Analysis lab


28. Aluminum hydroxide (Al(OH )3) is often present in antacids to neutralize stomach acid (HCl). The reaction occurs as follows: Al(OH)3(s) + 3HCl(aq) → AlC l3(aq) + 3 H2O(l). If 14.0 g of Al(OH )3

is present in an antacid tablet, determine the theoretical yield of AlCl3 produced when the tablet reacts with HCl.

29. Zinc reacts with iodine in a synthesis reaction: Zn + I2→ Zn I2.

a. Determine the theoretical yield if 1.912 mol of zinc is used.

b. Determine the percent yield if 515.6 g of product is recovered.30. Challenge When copper wire is placed into a silver nitrate solution

(AgNO3), silver crystals and copper(II) nitrate (Cu(N O3)2) solution form.

a. Write the balanced chemical equation for the reaction. b. If a 20.0-g sample of copper is used, determine the theoretical yield

of silver. c. If 60.0 g of silver is recovered from the reaction, determine the

percent yield of the reaction.

Based on Real Dat a 1, 2

Analyze and Conclude

Can rocks on the Moon provide an effective oxygen source for future lunar missions? Although the Moon has no atmosphere and thus no oxygen, its surface is covered with rocks and soil made from oxides. Scientists, looking for an oxygen source for future long-duration lunar missions, are researching ways to extract oxygen from lunar soil and rock. Analysis of samples col-lected during previous lunar missions provided scientists with the data shown in the table. The table identifies the oxides in lunar soil as well as each oxide’s percent-by-weight of the soil.

Think Critically1. Calculate For each of the oxides listed in the

table, determine the mass (in grams) that would exist in 1.00 kg of lunar soil.

2. Apply Scientists want to release the oxygen from its metal oxide using a decomposition reaction: metal oxide → metal + oxygen. To assess the viability of this idea, determine the amount of oxygen per kilogram contained in each of the oxides found in lunar soil.

3. Identify What oxide would yield the most oxygen per kilogram? The least?

4. Determine the theoretical yield of oxygen from the oxides present in a 1.00-kg sample of lunar soil.

Data and Observations

Moon-Rock Dat a 1

Oxide % Weight of Soil

Si O 2 47.3%

A l 2 O 3 17.8%

CaO 11.4%

FeO 10.5%

MgO 9.6%

Ti O 2 1.6%

N a 2 O 0.7%

K 2 O 0.6%

C r 2 O 3 0.2%

MnO 0.1%

1 Data obtained from: McKay, et al. 1994. JSC-1: A new lunar soil stimulant. Engineering, Construction, and Operations in Space IV: 857–866, American Society of Civil Engineers. 2 Data obtained from: Berggren, et al. 2005. Carbon monoxide silicate reduction system. Space Resources Roundtable VII.

5. Calculate Using methods currently available, scientists can produce 15 kg of oxygen from 100 kg of lunar soil. What is the percent yield of the process.


Section 11.4 • Percent Yield 387


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See Overset

AssessmentAssessmentPerformance Have each

student write four multiple choice questions involving percent yield calculations. Have them exchange their questions with a classmate and each answer the other’s questions. Have them resolve any disagree-ments about the correct answer. OL

3 AssessCheck for UnderstandingWrite the following equation on the board: CH 4 + 2 O 2 → C O 2 + 2 H 2 O. Ask students to determine the per-cent yield if 10.0 g of methane is burned producing 19.5 g of water. 86.7% yield OL

ReteachDiscuss the burning of gasoline in an internal combustion engine.2 C 8 H 18 + 2 5O 2 → 16C O 2 + 18 H 2 OAsk students to determine the percent yield of carbon dioxide if 700.0 g of octane produces 1800.0 g of carbon dioxide. Have students explain each step of the calculation. 83.4% yield CO 2 OL

ExtensionHave students research internal combustion engines and what is being done to increase the efficiency of automobiles and reduce air pollution. OL

Section 11.411.4 Assessment 31. percent yield 32. Not all reactions go to completion. Some of the reactants or products stick

to the surface of the container and are not massed or transferred. Other unexpected products form from competing reactions.

33. Divide the actual yield divided by the theoretical yield and multiply the quotient by 100.

34. 155.9 g Fe 2S 3 35. 87.9% yield of MgO

Section 11.411.4 Assessment

■ Figure 11.9 Sulfur, such as these piles at Vancouver Harbor, can be extract-ed from petroleum products by a chemical process. Sulfur is also mined by forcing hot water into underground deposits and pumping the liquid sulfur to the surface.

Percent Yield in the MarketplacePercent yield is important in the cost effectiveness of many industrial manufacturing processes. For example, the sulfur shown in Figure 11.9 is used to make sulfuric acid ( H 2 S O 4 ). Sulfuric acid is an important chemical because it is a raw material used to make products such as fertilizers, detergents, pigments, and textiles. The cost of sulfuric acid affects the cost of many of the consumer items you use every day. The first two steps in the manufacturing process are shown below.

Step 1 S 8 (s) + 8 O 2 (g) → 8S O 2 (g)Step 2 2S O 2 (g) + O 2 (g) → 2S O 3 (g)

In the final step, S O 3 combines with water to produce H 2 S O 4 .

Step 3 S O 3 (g) + H 2 O(l) → H 2 S O 4 (aq)

The first step, the combustion of sulfur, produces an almost 100% yield. The second step also produces a high yield if a catalyst is used at the relatively low temperature of 400°C. A catalyst is a substance that speeds a reaction but does not appear in the chemical equation. Under these conditions, the reaction is slow. Raising the temperature increases the reaction rate but decreases the yield.

To maximize yield and minimize time in the second step, engineers have devised a system in which the reactants, O 2 and S O 2 , are passed over a catalyst at 400°C. Because the reaction releases a great deal of heat, the temperature gradually increases with an accompanying decrease in yield. Thus, when the temperature reaches approximately 600°C, the mixture is cooled and then passed over the catalyst again. A total of four passes over the catalyst with cooling between passes results in a yield greater than 98%.

Section Summary◗ ◗ The theoretical yield of a chemical

reaction is the maximum amount of product that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation.

◗ ◗ The actual yield is the amount of product produced. Actual yield must be obtained through experimentation.

◗ ◗ Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is impor-tant in reducing the cost of every product produced through chemical processes.

31. MAIN Idea Identify which type of yield—theoretical yield, actual yield, or percent yield—is a measure of the efficiency of a chemical reaction.

32. List several reasons why the actual yield from a chemical reaction is not usually equal to the theoretical yield.

33. Explain how percent yield is calculated.

34. Apply In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(III) sulfide.

2Fe(s) + 3S(s) → F e 2 S 3 (s)

What is the theoretical yield, in grams, of iron(III) sulfide?

35. Calculate the percent yield of the reaction of magnesium with excess oxygen: 2Mg(s) + O 2 (g) → 2MgO(s)

Reaction Data

Mass of empty crucible 35.67 g

Mass of crucible and Mg 38.06 g

Mass of crucible and MgO (after heating) 39.15 g

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PurposeStudents will learn how the adapt-ability of HIV can be used to learn how new HIV drugs work.

BackgroundDrug resistance is conferred by mutations of HIV’s genetic sequence. HIV does not possess DNA (deoxy-ribonucleic acid), but rather stores its genetic information on a related molecule called RNA (ribonucleic acid). Researchers map the RNA sequence of HIV to discover which part of the RNA controls each part of the virus. When mutations are dis-covered, researchers can go back to that map to determine which part of the virus is altered by the mutation.

Teaching Strategies• Discuss examples of mutations that

provide resistance. These might include pesticide resistance in insects, herbicide resistance in weeds, and antibiotic resistance in disease-causing bacteria.

• Researchers encouraged resistance to PA-457 by administering low doses of the drug. Why not high doses? High doses might kill those viruses with only partial resistance. Compare this with the need to complete full courses of antibiotics to prevent the development of resistant bacteria.

• Note the crucial difference between viruses and bacteria: viruses are not affected by antibiot-ics. Have students find out why.

Chemistry


Resarch Answers will vary. Students should find that the dose-response relationship for therapeutic drugs var-ies. Some drugs are very predictable, requiring little moni-toring, but others require much closer monitoring. Disease, the patient’s age, and drug interactions can all affect the level of a drug in the patient’s system. Additionally, some drugs become toxic even at concentrations close to the therapeutic dose and need to be monitored very closely.

Battling Resistant StrainsThe human immunodeficiency virus (HIV), the virus that causes AIDS, has proven to be among the most incurable foes ever faced by modern medical science. One reason for this is the virus’s remarkable ability to adapt. Resistant strains of the virus appear quickly, rendering obsolete the newest and most powerful AIDS drugs. Now some researchers are using the virus’s adaptability as a way to fight it.

Selecting resistance PA-457 is a promising new anti-HIV drug synthesized from betulinic acid, an organic compound derived from some plants, including the bark of birch trees. To find out just what PA-457 does to HIV, known as the drug’s mechanism of action, researchers took what might seem a strange step: they encouraged samples of HIV to develop resistance to PA-457. Researchers subjected HIV samples to small doses of PA-457. Using a low dose made it more likely that some of the virus would survive the treatment and possibly develop resistance. Those viruses that survived exposure were collected, and their genetic sequences were examined. The surviving viruses were found to have a mutation in the genes that control how the virus builds a structure called a capsid, shown in Figure 1.

Figure 1 In a normal HIV virus, the capsid forms a protective coating around the genetic material.

Chemistry Research how scientists determine the safe dos-ing level for an experimental drug. Discuss how a drug’s effectiveness must be balanced with its potential toxicity and side effects. For more infor-mation on how a therapeutic dose is determined, visit glencoe.com.

Infectedhuman cell

Virusbudding out

Protein shielddestroyed

DefectiveRNA core

Capsid proteinshield forviral RNA

RNA core

Normal

With PA-457

Infectious Virus

“Dead” Virus

Figure 2 When treated with PA-457, the HIV capsid becomes misshapen and collapses, resulting in the death of the virus.

Surprise attack This finding was surprising, because it showed that, unlike most drugs, PA-457 attacks the HIV structure, rather than the enzymes that help HIV reproduce, as illustrated in Figure 2. This makes PA-457 among the first of a new class of HIV drugs known as maturation inhibitors—drugs that can prevent the virus from maturing during the late stages in its development.

Slowing evolution The hope is that because PA-457 and other maturation inhibitors attack the HIV structure, resistance will be slower to develop. Even so, maturation inhibitors will likely be prescribed in combination with other AIDS drugs that attack HIV at different stages of its life cycle. This practice, called multidrug therapy, makes it harder for HIV to develop resistance because any surviving virus would need to have multiple mutations—at least one for each anti-HIV drug. These mutations are less likely to occur at the same time.

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Chemistry & Health 389

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See the ChemLab worksheet in your FAST FILE.


PreparationTime Allotment one class period

Process Skills observe and infer, measure, collect data, interpret data, use numbers, compare and contrast, apply concepts

Safety Precautions Approve lab safety forms before work begins. Review the MSDS for copper(II) sul-fate pentahydrate and iron with stu-dents before beginning the lab. Copper(II) sulfate is toxic. Use only a GFCI outlet for the hot plate. Be sure students wear aprons, goggles, and plastic gloves. Students must wash their hands with soap after the lab.

Disposal The copper(II) sulfate and iron(II) sulfate solutions can be evaporated to dryness in the fume hood. The solid can then be disposed of at a designated waste facility.

Procedure• Decanting might be a new lab

technique for students. Show them how to slowly pour the liquid from the beaker down the stirring rod and into another beaker.

• Tell students that the copper metal will not stick to the stirring rod when it is completely dry.

• Troubleshooting For best results, use iron filings that are free of oxidation.

Analyze and ConcludeNumerical answers based on sample data. 1. Fe(s) + CuS O 4 (aq) → Cu(s) +

FeS O 4 (aq); 2.30 g Cu 2. 2.26 g Cu ; 0.0356 mol Cu; 0.0362 mol Fe;

Fe:Cu mole ratio = 1.02:1; percent yield = 98.3%

3. The ratio of Fe to Cu in the equation is 1:1, almost the same as the experimental ratio.

4. Copper was not completely dry; some copper could have oxidized if heated too much; copper could have been lost.

Inquiry ExtensionStudent answers will vary but should discuss possible error sources and their impact on the results.


Probeware Alternate CBL instructions can be found at glencoe.com.

INQUIRY EXTENSIONCompare your results with those of several other lab teams. Create a hypothesis to explain any differences.

DETERMINE THE MOLE RATIO

Background: Iron reacts with copper(II) sulfate (CuS O 4 ). By measuring the mass of iron that reacts and the mass of copper metal produced, you can calculate the experimental mole ratio.

Question: How does the experimental mole ratio compare with the theoretical mole ratio?

Materialscopper(II) sulfate penta- hot plate hydrate (CuS O 4 ·5 H 2 O) beaker tongsiron metal filings (20 mesh) balancedistilled water stirring rod150-mL beaker 400-mL beaker100-mL graduated cylinder weighing paper

Safety Precautions WARNING: Hot plates can cause burns. Turn off hot plates when not in use. Use only GFCI-protected circuits.

Procedure 1. Read and complete the lab safety form. 2. Measure the mass of a clean, dry 150-mL beaker.

Record all measurements in a data table. 3. Place approximately 12 g CuS O 4·5 H 2 O into the

150-mL beaker, and measure the combined mass. 4. Add 50 mL of distilled water to the CuS O 4 ·5 H 2 O.

Place the mixture on a hot plate set at medium, and stir until all of the solid dissolves (do not boil). Using tongs, remove the beaker from the hot plate.

5. Measure about 2 g of iron filings onto a piece of weighing paper. Measure the mass of the filings.

6. While stirring, slowly add the iron filings to the hot copper(II) sulfate solution. Be careful not to splash the hot solution.

7. Allow the reaction mixture to sit for 5 min. 8. Use the stirring rod to decant (pour off) the liquid

into a 400-mL beaker. Be careful to decant only the liquid—leave the solid copper metal behind.

9. Add 15 mL of distilled water to the copper solid, and carefully swirl the beaker to wash the copper. Decant the liquid into the 400-mL beaker.

10. Repeat Step 9 two more times. 11. Place the beaker containing the wet copper on the

hot plate. Use low heat to dry the copper.

12. After the copper is dry, use tongs to remove the beaker from the hot plate and allow it to cool.

13. Measure the mass of the beaker and the copper. 14. Cleanup and Disposal The dry copper can be

placed in a waste container. Moisten any residue that sticks to the beaker, and wipe it out using a paper towel. Pour the unreacted copper(II) sulfate and iron(II) sulfate solutions into a large beaker. Return all lab equipment to its proper place.

Analyze and Conclude 1. Apply Write a balanced chemical equation for the

reaction and calculate the mass of copper (Cu) that should have formed from the sample of iron (Fe) used. This mass is the theoretical yield.

2. Interpret Data Using your data, determine the mass and the moles of copper produced. Calculate the moles of iron used, and determine the whole-number iron-to-copper mole ratio and percent yield.

3. Compare and Contrast Compare the theoretical iron-to-copper mole ratio to the mole ratio you cal-culated using the experimental data.

4. Error Analysis Identify sources of the error that resulted in deviation from the mole ratio given in the balanced chemical equation.

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390 ChemLab

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Vocabulary PuzzleMaker glencoe.com

Study Guide

Use the VocabularyTo reinforce chapter vocabulary, have students write a sentence using each term. OL EL

Review Strategies• Have students explain why a bal-

anced chemical equation is neces-sary in stoichiometry. OL

• Have students list the steps in stoi-chiometric problem solving. OL

• Have students explain how a limit-ing reactant affects a stoichiomet-ric problem. OL

• Have students explain the percent yield of a reaction. OL

• Problems from p. 982 or the Supplemental Problems booklet can be used for review. OL

Use the ExamView® Assessment SuiteCD-ROM to:• create multiple versions of tests• create modified tests with one

mouse click• edit existing questions and add

your own questions• build tests aligned with state stan-

dards using built-in state curricu-lum tags

• change English tests to Spanish with one mouse click

• track students’ progress using the Teacher Management System

Students can visit glencoe.com to:• study the entire chapter online• access Web links for more informa-

tion, projects, and activities• review content online with the

Interactive Tutor and take Self-Check Quizzes

• take Chapter Tests and Standardized Test Practice

• use Study to Go to download con-tent onto a PDA

Vocabulary PuzzleMaker

For additional practice with vocabulary, have students access the Vocabulary PuzzleMaker online at glencoe.com.

Download quizzes, key terms, and flash cards from glencoe.com.

BIG Idea Mass relationships in chemical reactions confirm the law of conservation of mass.

Section 11.1Section 11.1 Defining Stoichiometry

MAIN Idea The amount of each reactant present at the start of a chemical reaction determines how much product can form.

Vocabulary• mole ratio (p. 371)• stoichiometry (p. 368)

Key Concepts• Balanced chemical equations can be interpreted in terms of moles, mass, and

representative particles (atoms, molecules, formula units).• The law of conservation of mass applies to all chemical reactions.• Mole ratios are derived from the coefficients of a balanced chemical equation.

Each mole ratio relates the number of moles of one reactant or product to the number of moles of another reactant or product in the chemical reaction.

Section 11.2Section 11.2 Stoichiometric Calculations

MAIN Idea The solution to every stoichiometric problem requires a balanced chemical equation.

Key Concepts• Chemists use stoichiometric calculations to predict the amounts of reactants

used and products formed in specific reactions.• The first step in solving stoichiometric problems is writing the balanced

chemical equation.• Mole ratios derived from the balanced chemical equation are used in

stoichiometric calculations.• Stoichiometric problems make use of mole ratios to convert between mass

and moles.

Section 11.3Section 11.3 Limiting Reactants

MAIN Idea A chemical reaction stops when one of the reactants is used up.

Vocabulary• excess reactant (p. 379)• limiting reactant (p. 379)

Key Concepts• The limiting reactant is the reactant that is completely consumed during a

chemical reaction. Reactants that remain after the reaction stops are called excess reactants.

• To determine the limiting reactant, the actual mole ratio of the available reactants must be compared with the ratio of the reactants obtained from the coefficients in the balanced chemical equation.

• Stoichiometric calculations must be based on the limiting reactant.

Section 11.4Section 11.4 Percent Yield

MAIN Idea Percent yield is a measure of the efficiency of a chemical reaction.

Vocabulary• actual yield (p. 385)• percent yield (p. 386)• theoretical yield (p. 385)

Key Concepts• The theoretical yield of a chemical reaction is the maximum amount of product

that can be produced from a given amount of reactant. Theoretical yield is calculated from the balanced chemical equation.

• The actual yield is the amount of product produced. Actual yield must be obtained through experimentation.

• Percent yield is the ratio of actual yield to theoretical yield expressed as a percent. High percent yield is important in reducing the cost of every product produced through chemical processes.

Percent yield = actual yield

__ theoretical yield × 100

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Chapter 11 • Study Guide 391

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Chapter Test glencoe.com

Assessment

Section 11.1

Mastering Concepts 36. Mole ratios are determined by the coeffi-

cients in a balanced equation. If the equation is not balanced, the relation-ship between reactants and products cannot be determined.

37. relationships among particles, moles, and mass for all reactants and products

38. Mole ratios allow for the conversion from moles of one substance in a balanced chemical equation to moles of another substance in the same equation.

39. moles B/moles A 40. The coefficients in the balanced chemical

equation show the numbers of represen-tative particles involved in a reaction. Subscripts give the numbers of different kinds of atoms within a molecule or for-mula unit.

41. The mass of the reactants will always equal the mass of the products.

42. 1 mol ( NH 4 ) 2 Cr 2 O 7 /1 mol N 2 and inverse; 1 mol ( NH 4 ) 2 Cr 2 O 7 /1 mol Cr 2 O 3 and inverse; 1 mol ( NH 4 ) 2 Cr 2 O 7 /4 mol H 2 O and inverse

43. 2 M 2 N → M 4 + N 2 ; 1 mol N 2 /2 mol M 2 N, 1 mol N 2 /1 M 42 mol M 2 N/1 mol M 4, 2 mol M 2 N/1 molN 2 1 mol M 4 /1 mol N 2 , 1 mol M 4 /2 mol M 2 N

Mastering Problems 44. 4 atoms Al + 3 molecules O 2 →

2 formula units Al 2 O 3 ; 4 mol Al + 3 mol O 2 → 2 mol Al 2 O 3 ; 107.93 g Al + 95.99 g O 2 → 203.92 g Al 2 O 3

45. 1 formula unit Sn O 2 + 2 atoms C → 1 atom Sn + 2 molecules CO; 1 mol Sn O 2 + 2 mol C → 1 mol Sn + 2 mol CO; 150.71 g Sn O 2 + 24.02 g C → 118.71 g Sn + 56.02 g CO

46. Cu(s) + 4HN O 3 (aq) → Cu ( NO 3 ) 2 (aq) + 2N O 2 (g) + 2 H 2 O(l); answers may include 1 mol Cu/4 mol HN O 3 and inverse,1 mol Cu/1 mol Cu ( NO 3 ) 2 and inverse1 mol Cu/2 mol NO 2 and inverse1 mol Cu/2 mol H 2 O and inverse4 mol HN O 3 /1 mol Cu ( NO 3 ) 2 and inverse4 mol HN O 3 /2 mol NO 2 and inverse4 mol HNO 3 /2 mol H 2 O and inverse1 mol Cu ( NO 3 ) 2 /2 mol NO 2 and inverse

1 mol Cu ( NO 3 ) 2 /2 mol H 2 O and inverse2 mol NO 2 /2 mol H 2 O and inverse

47. a. 2HCl(aq) + Pb ( NO 3 ) 2 (aq) → PbCl 2 (s) + 2HNO 3 (aq)

b. 2 molecules HCl + 1 formula unit Pb ( NO 3 ) 2 → 1 formula unit PbCl 2 + 2 molecules HNO 3 ; 2 mol HCl + 1 mol Pb ( NO 3 ) 2 → 1 mol PbCl 2 + 2 mol HNO 3 ; 72.9 g HCl + 331.2 g Pb ( NO 3 ) 2 → 278.1 g PbCl 2 + 126.0 g HNO 3

48. 2 mol Fe/1 mol Fe 2 O 3 49. a. Si O 2 (s) + 4HF(aq) → SiF 4 (g) + 2 H 2 O(l)

b. Students may write any 3 of the 12 ratios. Explanations should correctly describe how the ratio can be used as a conversion factor.

50. 1 mol FeCr 2 /1 mol FeCr 2 O 4 51. 2 mol CaSO 4 /2 mol SO 2 52. 3W + X → 2Y + 4Z 53. a. 1Mg (OH) 2 + 2HCl → 1MgCl 2 + 2H 2 O b. 1 mol MgCl 2 /1 mol Mg (OH) 2

or 1 mol MgCl 2 /2 mol HCl

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Section 11.1Mastering Concepts

36. Why must a chemical equation be balanced before you can determine mole ratios?

37. What relationships can be determined from a balanced chemical equation?

38. Explain why mole ratios are central to stoichiometric calculations.

39. What is the mole ratio that can convert from moles of A to moles of B?

40. Why are coefficients used in mole ratios instead of subscripts?

41. Explain how the conservation of mass allows you to interpret a balanced chemical equation in terms of mass.

42. When heated by a flame, ammonium dichromate decomposes, producing nitrogen gas, solid chromium(III) oxide, and water vapor. (N H 4 )2C r 2 O 7 → N 2 + C r 2 O 3 + 4 H 2 O

Write the mole ratios for this reaction that relate ammo-nium dichromate to the products.

■ Figure 11.10

43. Figure 11.10 depicts an equation with squares repre-senting Element M and circles representing Element N. Write a balanced equation to represent the picture shown, using smallest whole-number ratios. Write mole ratios for this equation.

Mastering Problems 44. Interpret the following equation in terms of particles,

moles, and mass. 4Al(s) + 3 O 2 (g) → 2A l 2 O 3 (s)

45. Smelting When tin(IV) oxide is heated with carbon in a process called smelting, the element tin can be extracted. Sn O 2 (s) + 2C(s) → Sn(l) + 2CO(g)

Interpret the chemical equation in terms of particles, moles, and mass.

46. When solid copper is added to nitric acid, copper(II) nitrate, nitrogen dioxide, and water are produced. Write the balanced chemical equation for the reaction. List six mole ratios for the reaction.

47. When hydrochloric acid solution reacts with lead(II) nitrate solution, lead(II) chloride precipitates and a solution of nitric acid is produced.

a. Write the balanced chemical equation for the reaction. b. Interpret the equation in terms of molecules and

formula units, moles, and mass. 48. When aluminum is mixed with iron(III) oxide, iron

metal and aluminum oxide are produced, along with a large quantity of heat. What mole ratio would you use to determine moles of Fe if moles of F e 2 O 3 is known?F e 2 O 3 (s) + 2Al(s) → 2Fe(s) + A l 2 O 3 (s) + heat

49. Solid silicon dioxide, often called silica, reacts with hydrofluoric acid (HF) solution to produce the gas silicon tetrafluoride and water.

a. Write the balanced chemical equation for the reaction. b. List three mole ratios, and explain how you would use

them in stoichiometric calculations. 50. Chrome The most important commercial ore of chro-

mium is chromite (FeC r 2 O 4 ). One of the steps in the process used to extract chromium from the ore is the reaction of chromite with coke (carbon) to produce ferrochrome (FeC r 2 ). 2C(s) + FeC r 2 O 4 (s) → FeC r 2 (s) + 2C O 2 (g)

What mole ratio would you use to convert from moles of chromite to moles of ferrochrome?

51. Air Pollution The pollutant S O 2 is removed from the air by in a reaction that also involves calcium carbonate and oxygen. The products of this reaction are calcium sulfate and carbon dioxide. Determine the mole ratio you would use to convert moles of S O 2 to moles of CaS O 4 .

52. Two substances, W and X, react to form the products Y and Z. Table 11.2 shows the moles of the reactants and products involved when the reaction was carried out. Use the data to determine the coefficients that will bal-ance the equation W + X → Y + Z.

Table 11.2 Reaction Data

Moles of Reactants Moles of Products

W X Y Z

0.90 0.30 0.60 1.20

53. Antacids Magnesium hydroxide is an ingredient in some antacids. Antacids react with excess hydrochloric acid in the stomach to relieve indigestion.___Mg(OH ) 2 + ___HCl → ___ MgC l 2 + ___ H 2 O

a. Balance the reaction of Mg (OH ) 2 with HCl. b. Write the mole ratio that would be used to determine

the number of moles of MgC l 2 produced when HCl reacts with Mg(OH ) 2 .

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392 Chapter 11 • Assessment


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70. 1.50 × 102 g/mol71. 34 L72. 2.93 g/L73. tk74. tk75. tk76. tk77. tk78. tk

Section 13.3

Section 11.2

Mastering Concepts 54. Write a balanced chemical equation for

the reaction. 55. The balanced equation provides the rela-

tionship between reactants and prod-ucts, and the coefficients in the equation are used to write mole ratios relating reactants and products.

56. Stoichiometry is based on the law of con-servation of mass. The calculations are used to determine the mass of reactants and products. Once found, the sum of reactants will equal the sum of products, verifying the law of conservation of mass.

57. Molar mass is a conversion factor for con-verting moles of a given substance to mass or mass of a given substance to moles.

58. You must have the balanced chemical equation and know the quantity of one substance in the reaction other than the product you are to determine.

59. a. 2H 2 S(g) + O 2 (g) → 2H 2 O(g) + 2S(s)

b. Student sketches should show the formation of six water molecules (H 2 O) and six sulfur atoms (S).

Mastering Problems 60. C 6 H 12 O 6 → 2 C 2 H 5 OH + 2 CO 2 ;

390 g C 2 H 5 OH 61. 5.50 mol of C 2 H 2 62. 0.0119 mol Na 3C 6H 5O 7 63. 523 g C 3 H 7 COO C 2 H 5 64. 2 C 8 H 18 (l) + 25 O 2 (g) →

16 CO 2 (g) + 18H 2 O(l); 71.4 g C 8 H 18

65. a. K 2 Cr O 4 (aq) + Pb ( NO 3 ) 2 (aq) →PbCr O 4 (s) + 2KN O 3 (aq)

b. 80.8 g PbCr O 4 66. a. N 2 H 2 (l) + H 2 O 2 (l) →

N 2 (g) + 2 H 2 O(g) b. 3.00 × 10 2 g N 2 H 2 67. 6.72 g CH 4 68. mass K O 2 = 1100 g, mass H 2 O = 140 g,

mass CO 2 = 7 × 102 g, mass KH CO 3 = 1600 g

69. C 2 H 5 OH(l) + 3 O 2 (g) → 2C O 2 (g) + 3 H 2 O(l); 191.0 g CO 2


54. What is the first step in all stoichiometric calculations? 55. What information does a balanced equation provide? 56. On what law is stoichometry based, and how do the

calculations support this law? 57. How is molar mass used in some stoichiometric

calculations? 58. What information must you have in order to calculate

the mass of product formed in a chemical reaction?

+■ Figure 11.11

59. Each box in Figure 11.11 represents the contents of a flask. One flask contains hydrogen sulfide, and the other contains oxygen. When the contents of the flasks are mixed, a reaction occurs and water vapor and sulfur are produced. In the figure, the red circles represent oxygen, the yellow circles represent sulfur, and blue circles repre-sent hydrogen.

a. Write the balanced chemical equation for the reaction. b. Using the same color code, sketch a representation of

the flask after the reaction occurs.

Mastering Problems 60. Ethanol ( C 2 H 5 OH), also known as grain alcohol, can

be made from the fermentation of sugar ( C 6 H 12 O 6 ). The unbalanced chemical equation for the reaction is shown below. ___ C 6 H 12 O 6 → ___ C 2 H 5 OH + ___C O 2

Balance the chemical equation and determine the mass of C 2 H 5 OH produced from 750 g of C 6 H 12 O 6 .

61. Welding If 5.50 mol of calcium carbide (Ca C 2 ) reacts with an excess of water, how many moles of acetylene ( C 2 H 2 ), a gas used in welding, will be produced?

Ca C 2 (s) + 2 H 2 O(l) → Ca(OH ) 2 (aq) + C 2 H 2 (g)

62. Antacid Fizz When an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate (NaHC O 3 ), also called sodium bicarbonate, and citric acid ( H 3 C 6 H 5 O 7 ).3NaHC O 3 (aq) + H 3 C 6 H 5 O 7 (aq) → 3C O 2 (g) + 3 H 2 O(l) + N a 3 C 6 H 5 O 7 (aq)

How many moles of N a 3 C 6 H 5 O 7 can be produced if one tablet containing 0.0119 mol of NaHC O 3 is dissolved?

63. Esterification The process in which an organic acid and an alcohol react to form an ester and water is known as esterification. Ethyl butanoate ( C 3 H 7 COO C 2 H 5 ), an ester, is formed when the alcohol ethanol ( C 2 H 5 OH) and butanoic acid ( C 3 H 7 COOH) and are heated in the presence of sulfuric acid. C 2 H 5 OH(l) + C 3 H 7 COOH(l) → C 3 H 7 COO C 2 H 5 (l) + H 2 O(l)

Determine the mass of ethyl butanoate produced if 4.50 mol of ethanol is used.

64. Greenhouse Gas Carbon dioxide is a greenhouse gas that is linked to global warming. It is released into the atmosphere through the combustion of octane ( C 8 H 18 ) in gasoline. Write the balanced chemical equation for the combustion of octane and calculate the mass of octane needed to release 5.00 mol of C O 2 .

65. A solution of potassium chromate reacts with a solution of lead(II) nitrate to produce a yellow precipitate of lead(II) chromate and a solution of potassium nitrate.

a. Write the balanced chemical equation. b. Starting with 0.250 mol of potassium chromate,

determine the mass of lead chromate formed. 66. Rocket Fuel The exothermic reaction between liquid

hydrazine ( N 2 H 2 ) and liquid hydrogen peroxide ( H 2 O 2 ) is used to fuel rockets. The products of this reaction are nitrogen gas and water.

a. Write the balanced chemical equation. b. How much hydrazine, in grams, is needed to produce

10.0 mol of nitrogen gas? 67. Chloroform (CHC l 3 ), an important solvent, is produced

by a reaction between methane and chlorine. C H 4 (g) + 3C l 2 (g) → CHC l 3 (g) + 3HCl(g)

How much C H 4 , in grams, is needed to produce 50.0 grams of CHC l 3 ?

68. Oxygen Production The Russian Space Agency uses potassium superoxide (K O 2 ) for the chemical oxygen generators in their space suits. 4K O 2 + 2 H 2 O + 4C O 2 → 4KHC O 3 + 3 O 2

Complete Table 11.3.

Table 11.3 Oxygen Generation Reaction Data

Mass K O 2

Mass H 2 O

Mass C O 2

Mass KHC O 3

Mass O 2

380 g

69. Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of C O 2 produced from the combustion of 100.0 g of ethanol. C 2 H 5 OH(l) + O 2 (g) → C O 2 (g) + H 2 O(g)

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Chapter 11 • Assessment 393


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70. a. Pb(s) + Pb O 2 (s) + 2 H 2 S O 4 (aq) → 2PbS O 4 (aq) + 2 H 2 O(l)

b. 73.2 g Pb SO 4 71. a. 50.2 g Au b. 33.5% gold in ore 72. 0.587 g Na 3 Ag ( S 2 O 3 ) 2

Section 11.3

Mastering Concepts 73. The actual mole ratio of reactants from

the chemical equation is compared to the mole ratio determined from the given quantities.

74. The limiting reactant is the reactant that produces the lowest number of moles of product. Mass does not determine the limiting reactant but the number of moles.

75. a. 3M2 + N2→ 2M3N b. 6 moles of element M (in the form of

3 moles of M2) and 6 moles of ele-ment N (likewise, 3 moles of N2)

c. 2 moles of M3N form with 2 moles of N2 unreacted (4 total moles of ele-ment N)

d. M2 is the limiting reactant and N2 is the excess reactant.

Mastering Problems 76. Hydrogen is limiting; ethyne is the excess

reactant. One mol of ethyne is left over. 77. 4.0 mol Fe(OH) 2 78. 10.0 mol CsXeF 7 79. 1120 g Fe 80. C l 2 is the limiting reactant; phosphorus is

in excess. 81. a. MnO 2 is the limiting reactant. b. 17.1 g Zn (OH) 2 82. 2Li(s) + Br2(l) → 2LiBr(s) a. Br 2 b. 27.1 g LiBr c. Li, 22.8 g

70. Car Battery Car batteries use lead, lead(IV) oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(II) sulfate in solu-tion and water.

a. Write the balanced equation for the reaction. b. Determine the mass of lead(II) sulfate produced

when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid.

71. To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water.4Au(s) + 8NaCN(aq) + O 2 (g) + 2 H 2 O(l) →

4NaAu(CN ) 2 (aq) + 4NaOH(aq) a. Determine the mass of gold that can be extracted if

25.0 g of sodium cyanide is used. b. If the mass of the ore from which the gold was

extracted is 150.0 g, what percentage of the ore is gold? 72. Film Photographic film contains silver bromide in gela-

tin. Once exposed, some of the silver bromide decom-poses, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (N a 3 Ag( S 2 O 3 ) 2 ) is produced.AgBr(s) + 2N a 2 S 2 O 3 (aq) → N a 3 Ag( S 2 O 3 ) 2 (aq) + NaBr(aq)

Determine the mass of N a 3 Ag( S 2 O 3 ) 2 produced if 0.275 g of AgBr is removed.


73. How is a mole ratio used to find the limiting reactant? 74. Explain why the statement, “The limiting reactant is the

reactant with the lowest mass” is incorrect.

■ Figure 11.12

75. Figure 11.12 uses squares to represent Element M and circles to represent Element N.

a. Write the balanced equation for the reaction. b. If each square represents 1 mol of M and each circle

represents 1 mol of N, how many moles of M and N were present at the start of the reaction?

c. How many moles of product form? How many moles of Element M and Element N are unreacted?

d. Identify the limiting reactant and the excess reactant.

Mastering Problems

Ethyne Hydrogen Ethane Ethyne

+ +→

■ Figure 11.13

76. The reaction between ethyne ( C 2 H 2 ) and hydrogen ( H 2 ) is illustrated in Figure 11.13. The product is ethane ( C 2 H 6 ). Which is the limiting reactant? Which is the excess reactant? Explain.

77. Nickel-Iron Battery In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery.Fe(s) + 2NiO(OH)(s) + 2 H 2 O(l) → Fe(OH ) 2 (s) + 2Ni(OH ) 2 (aq)How many mol of Fe(OH ) 2 is produced when 5.00 mol of Fe and 8.00 mol of NiO(OH) react?

78. One of the few xenon compounds that form is cesium xenon heptafluoride (CsXe F 7 ). How many moles of CsXe F 7 can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride? CsF(s) + Xe F 6 (s) → CsXe F 7 (s)

79. Iron Production Iron is obtained commercially by the reaction of hematite (F e 2 O 3 ) with carbon monoxide. How many grams of iron is produced when 25.0 mol of hematite reacts with 30.0 mol of carbon monoxide? F e 2 O 3 (s) + 3CO(g) → 2Fe(s) + 3C O 2 (g)

80. The reaction of chlorine gas with solid phosphorus ( P 4 ) produces solid phosphorus pentachloride. When 16.0 g of chlorine reacts with 23.0 g of P 4 , which reactant is limiting? Which reactant is in excess?

81. Alkaline Battery An alkaline battery produces electrical energy according to this equation.Zn(s) + 2Mn O 2 (s) + H 2 O(l) → Zn(OH ) 2 (s) + M n 2 O 3 (s)

a. Determine the limiting reactant if 25.0 g of Zn and 30.0 g of Mn O 2 are used.

b. Determine the mass of Zn(OH ) 2 produced. 82. Lithium reacts spontaneously with bromine to produce

lithium bromide. Write the balanced chemical equation for the reaction. If 25.0 g of lithium and 25.0 g of bromine are present at the beginning of the reaction, determine

a. the limiting reactant. b. the mass of lithium bromide produced. c. the excess reactant and the excess mass.

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Section 11.4Mastering Concepts 83. Actual yield is the amount of product

obtained experimentally. Theoretical yield is the amount of product predicted by a stoichiometric calculation.

84. Actual yield is determined through experimentation. Theoretical yield is cal-culated from a given reactant or the lim-iting reactant.

85. No, you cannot produce more product than the theoretical yield, which is deter-mined from the starting reactants.

86. (actual yield/theoretical yield) × 100 = percent yield

87. The quantity of one reactant and the actual yield of the product .

88. The mass of one substance in the reac-tion and the actual mass of metal hydroxide produced.

89. The reaction did not go to completion. Using squares to represent Element A and circles to represent Element B, the initial products would have yielded four AB 2 particles, but only three were pro-duced. There are enough remaining unreacted A and B particles to produce one more AB 2 particle. The percent yield is 75%.

Mastering Problems 90. theoretical yield = 369 g C 2 H 5 OH

percent yield C 2 H 5 OH = 94.6% 91. a. 2PbS + 3 O 2 → 2PbO + 2 SO 2

theoretical yield = 186.6 g PbO b. percent yield PbO = 91.10% 92. a. theoretical yield CO 2 = 103 g b. percent yield CO 2 = 94.7% 93. a. HF is the limiting reactant. b. SiO 2 is the excess reactant, 20.0 g. c. theoretical yield = 48.0 g H 2 Si F 6 d. percent yield H 2 Si F 6 = 95.4% 94. 73.0% yield Zr 95. mass CH 3 OH(l) = 9.71 g moles CO(g) = 0.303 mol moles CH 3 OH(l) = 0.303 mol percent yield = 87.7% 96. theoretical yield = 49.92 g P4

percent yield P 4 = 90.1% 97. theoretical yield = 24.3 g Cl 2 percent yield C l 2 = 82.3%

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83. What is the difference between actual yield and theoreti-cal yield?

84. How are actual yield and theoretical yield determined? 85. Can the percent yield of a chemical reaction be more

than 100%? Explain your answer. 86. What relationship is used to determine the percent yield

of a chemical reaction? 87. What experimental information do you need in order to

calculate both the theoretical and the percent yield of any chemical reaction?

88. A metal oxide reacts with water to produce a metal hydroxide. What additional information would you need to determine the percent yield of metal hydroxide from this reaction?

Element AElement B

■ Figure 11.14

89. Examine the reaction represented in Figure 11.14. Determine if the reaction went to completion. Explain your answer, and calculate the percent yield of the reaction.

Mastering Problems 90. Ethanol ( C 2 H 5 OH) is produced from the fermentation

of sucrose ( C 12 H 22 O 11 ) in the presence of enzymes. C 12 H 22 O 11 (aq) + H 2 O(g) → 4 C 2 H 5 OH(l) + 4C O 2 (g) Determine the theoretical yield and the percent yield of

ethanol if 684 g of sucrose undergoes fermentation and 349 g of ethanol is obtained.

91. Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is: PbS(s) + O 2 (g) → PbO(s) + S O 2 (g)

a. Balance the equation, and determine the theoretical yield of PbO if 200.0 g of PbS is heated.

b. What is the percent yield if 170.0 g of PbO is obtained? 92. Upon heating, calcium carbonate (CaC O 3 ) decomposes

to calcium oxide (CaO) and carbon dioxide (C O 2 ). a. Determine the theoretical yield of C O 2 if 235.0 g of

CaC O 3 is heated. b. What is the percent yield of C O 2 if 97.5 g of C O 2 is

collected?

93. Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosilicic acid ( H 2 Si F 6 ).

Si O 2 (s) + 6HF(aq) → H 2 Si F 6 (aq) + 2 H 2 O(l) 40.0 g Si O 2 and 40.0 g HF react to yield 45.8 g H 2 Si F 6 .

a. What is the limiting reactant? b. What is the mass of the excess reactant? c. What is the theoretical yield of H 2 Si F 6 ? d. What is the percent yield? 94. Van Arkel Process Pure zirconium is obtained using

the two-step Van Arkel process. In the first step, impure zirconium and iodine are heated to produce zirconium iodide (Zr I 4 ). In the second step, Zr I 4 is decomposed to produce pure zirconium. Zr I 4 (s) → Zr(s) + 2 I 2 (g)

Determine the percent yield of zirconium if 45.0 g of Zr I 4 is decomposed and 5.00 g of pure Zr is obtained.

95. Methanol, wood alcohol, is produced when carbon monoxide reacts with hydrogen gas. CO + 2 H 2 → C H 3 OH

When 8.50 g of carbon monoxide reacts with an excess of hydrogen, 8.52 g of methanol is collected. Complete Table 11.4, and calculate the percent yield.

Table 11.4 Methanol Reaction Data

CO(g) C H 3 OH(l)

Mass 8.52 g

Molar mass 28.01 g/mol 32.05 g/mol

Moles

96. Phosphorus ( P 4 ) is commercially prepared by heating a mixture of calcium phosphate (CaSi O 3 ), sand (Si O 2 ), and coke (C) in an electric furnace. The process involves two reactions.2C a 3 (P O 4 ) 2 (s) + 6Si O 2 (s) → 6CaSi O 3 (l) + P 4 O 10 (g) P 4 O 10 (g) + 10C(s) → P 4 (g) + 10CO(g)

The P 4 O 10 produced in the first reaction reacts with an excess of coke (C) in the second reaction. Determine the theoretical yield of P 4 if 250.0 g of C a 3 (P O 4 ) 2 and 400.0 g of Si O 2 are heated. If the actual yield of P 4 is 45.0 g, determine the percent yield of P 4 .

97. Chlorine forms from the reaction of hydrochloric acid with manganese(IV) oxide. The balanced equation is: Mn O 2 + 4HCl → MnC l 2 + C l 2 + 2 H 2 O

Calculate the theoretical yield and the percent yield of chlorine if 86.0 g of Mn O 2 and 50.0 g of HCl react. The actual yield of C l 2 is 20.0 g.

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Mixed Review 98. 2 mol NH 4 N O 3 /1 mol CuS 99. 401 g CaNCN 100. 25.6 g Cu 101. a. 2NO(g) + O 2 (g) → 2N O 2 (g) b. 2 mol NO 2 /2 mol NO 102. theoretical yield = 4.04 g H 2 percent yield H 2 = 94.1% 103. The graph levels off because the oxy-

gen limits the reaction at that point. fixed amount = 0.329 mol O 2

Think Critically 104. No, percent yields cannot be greater

than 100%. High results could mean the product was not completely dry, or it was contaminated.

105. a. No, because there is only one reactant.

b. Yes, because there are two reac-tants. Not enough information is given to identify which is the limit-ing reactant.

106. Obtain and record the mass of an empty evaporating dish. Add 2.00 g of copper(II) sulfate pentahydrate and obtain and record the mass of the hydrate and evaporating dish. Heat the dish gently for 5 minutes, then strongly for 5 minutes to drive off the water. Cool the dish and remeasure the mass. Record. Determine the mass of the anhydrous copper sulfate. Using the equation Cu SO 4 •5 H 2 O → Cu SO 4 + 5 H 2 O and the initial mass of the copper(II) sulfate pentahydrate, deter-mine the theoretical yield of copper(II) sulfate. Determine the actual yield of copper(II) sulfate. Divide the actual yield by the theoretical yield and multi-ply by 100 to determine percent yield of copper(II) sulfate.

107. When you fan the flame, additional oxygen is added and the remaining coals can burn.

108. a. 2 Na 3 P O 4 (aq) + 3Co ( NO 3 ) 2 (aq) → Co 3 ( PO 4 ) 2 (s) + 6Na NO 3 (aq)

b. Trial 1: limiting reactant is Na 3 PO 4 , excess reactant is Co ( NO 3 ) 2 .Trials 2 through 4: limiting reactant is Co ( NO 3 ) 2 , excess reactant is Na 3 PO 4 .

Mixed Review 98. Ammonium sulfide reacts with copper(II) nitrate in a

double replacement reaction. What mole ratio would you use to determine the moles of N H 4 N O 3 produced if the moles of CuS are known?

99. Fertilizer The compound calcium cyanamide (CaNCN) is used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures. Ca C 2 (s) + N 2 (g) → CaNCN(s) + C(s)

What mass of CaNCN can be produced if 7.50 mol of Ca C 2 reacts with 5.00 mol of N 2 ?

100. When copper(II) oxide is heated in the presence of hydrogen gas, elemental copper and water are produced. What mass of copper can be obtained if 32.0 g of copper(II) oxide is used?

101. Air Pollution Nitrogen oxide, which is present in urban air pollution, immediately converts to nitrogen dioxide as it reacts with oxygen.

a. Write the balanced chemical equation for the forma-tion of nitrogen dioxide from nitrogen oxide.

b. What mole ratio would you use to convert from moles of nitrogen oxide to moles of nitrogen dioxide?

102. Electrolysis Determine the theoretical and percent yield of hydrogen gas if 36.0 g of water undergoes elec-trolysis to produce hydrogen and oxygen and 3.80 g of hydrogen is collected.

500

10 15 25 30 3520

Mas

s of

Fe 2

O3

(g)

30

20

10

Mass of Fe (g)

Mass of Fe2O3 Formed From Burning Fe

■ Figure 11.15

103. Iron reacts with oxygen as shown. 4Fe(s) + 3 O 2 (g) → 2F e 2 O 3 (s)

Different amounts of iron were burned in a fixed amount of oxygen. For each mass of iron burned, the mass of iron(II) oxide formed was plotted on the graph shown in Figure 11.15. Why does the graph level off after 25.0 g of iron is burned? How many moles of oxygen are present in the fixed amount?

Think Critically

104. Analyze and Conclude In an experiment, you obtain a percent yield of product of 108%. Is such a percent yield possible? Explain. Assuming that your calculation is correct, what reasons might explain such a result?

105. Observe and Infer Determine whether each reaction depends on a limiting reactant. Explain why or why not, and identify the limiting reactant.

a. Potassium chlorate decomposes to form potassium chloride and oxygen.

b. Silver nitrate and hydrochloric acid react to produce silver chloride and nitric acid.

106. Design an Experiment Design an experiment that can be used to determine the percent yield of anhydrous copper(II) sulfate when copper(II) sulfate pentahydrate is heated to remove water.

107. Apply When a campfire begins to die down and smolder, you can rekindle the flame by fanning the fire. Explain, in terms of stoichiometry, why the fire again begins to flare up when fanned.

108. Apply Students conducted a lab to investigate limiting and excess reactants. The students added different volumes of sodium phosphate solution (N a 3 P O 4 ) to a beaker. They then added a constant volume of cobalt(II) nitrate solution (Co(N O 3 ) 2 ), stirred the contents, and allowed the beakers to sit overnight. The next day, each beaker had a purple precipitate at the bottom. The stu-dents decanted the supernatant from each beaker, divid-ed it into two samples, and added one drop of sodium phosphate solution to one sample and one drop of cobalt(II) nitrate solution to the second sample. Their results are shown in Table 11.5.

a. Write a balanced chemical equation for the reaction. b. Based on the results, identify the limiting reactant

and the excess reactant for each trial.

Table 11.5 Reaction Data for Co(N O 3 ) 2 and N a 3 P O 4

TrialVolume N a 3 P O 4

Volume Co(N O 3 ) 2

Reaction with Drop of

N a 3 P O 4

Reaction with Drop

of Co(N O 3 ) 2

1 5.0 mL 10.0 mLpurple

precipitateno reaction

2 10.0 mL 10.0 mL no reactionpurple

precipitate


precipitate


precipitate

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Challenge Problem

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Challenge Problem 109. a. V 2 O 5 and V O 2 b. V 2 O 5 + H 2 → 2V O 2 + H 2 O

V O 2 + 2H 2 → V + 2H 2 O c. 0.532 g H 2

Cumulative Review 110. a. hypothesis, because it is based only

on observation, not on data 111. a. [He]2 s 2 2 p 5 b. [Ne]3 s 2 3 p 1 c. [Ar]4 s 2 3 d 2 d. [Xe]6 s 2 4 f 14 5 d 10 6 p 6 112. Diatomic molecules achieve a noble

gas electron configuration by forming covalent bonds. The monoatomic gases already have noble gas electron configurations.

113. 4K(s) + O 2 (g) → 2 K 2 O(s) 114. 352.02 amu; 352.02 g/mol 115. a. They are related by a whole-

number multiplier. b. C 2 H 4 O

Additional Assessment

Chemistry


116. Answers will vary. Common pollutants are S O 2 , NO, N O 2 and O 3 . Check the stoichiometry and be sure it accounts for a decrease in the pollutant.

117. Answers will vary. Make sure the fol-lowing equation is included. N 2 (g) + 3 H 2 (g) → 2N H 3 (g) + 92 kJThe aim of the Haber process was to control a reaction so that a large amount of a useful product was yielded quickly. The process was of great importance because the Germans had to come up with a nitrogen compound that could be produced in large amounts.

Document-Based QuestionsData obtained from: Becker, Bob. April 2006. ChemMatters. 24: no. 2.

118. H 2 O 2 is the limiting reactant. 119. 19.1 mg of C6H4(OH)2 excess 120. 79.4 mg C 6 H 4 O 2

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Additional Assessment

Chemistry

116. Air Pollution Research the air pollutants produced by combustion of gasoline in internal combustion engines. Discuss the common pollutants and the reaction that produces them. Show, through the use of stoichiometry, how each pollutant could be reduced if more people used mass transit.

117. Haber Process The percent yield of ammonia pro-duced when hydrogen and nitrogen are combined under ordinary conditions is extremely small. However, the Haber Process combines the two gases under a set of conditions designed to maximize yield. Research the conditions used in the Haber Process, and find out why the development of the process was of great importance.

Document-Based QuestionChemical Defense Many insects secrete hydrogen peroxide ( H 2 O 2 ) and hydroquinone C 6 H 4 (OH ) 2 . Bombardier beetles take this a step further by mixing these chemicals with a cat-alyst. The result is an exothermic chemical reaction and a spray of hot, irritating chemicals for any would-be predator. Researchers hope to use a similar method to reignite aircraft turbine engines.

Figure 11.17 below shows the unbalanced chemical reaction that results in the bombardier beetle’s defensive spray.

Data obtained from: Becker, Bob. April 2006. ChemMatters. 24: no. 2.

Hydroquinone

H2O2

C6H4O2C6H4(OH)2

H2O O2 Energy

OH

OH

O

O

+ + + +

Benzoquinone

Catalyst

■ Figure 11.17

118. Balance the equation in Figure 11.17. If the bombardier beetle stores 100.0 mg of hydroquinone ( C 6 H 4 (OH ) 2 ) along with 50.0 mg of hydrogen perox-ide ( H 2 O 2 ), what is the limiting reactant?

119. What is the excess reactant and how many milligrams are in excess?

120. How many milligrams of benzoquinone will be produced?

Challenge ProblemChallenge Problem 109. When 9.59 g of a certain vanadium oxide is heated in

the presence of hydrogen, water and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. When the second vanadium oxide undergoes additional heating in the presence of hydro-gen, 5.38 g of vanadium metal forms.

a. Determine the empirical formulas for the two vanadium oxides.

b. Write balanced equations for the steps of the reaction. c. Determine the mass of hydrogen needed to complete

the steps of this reaction.

Cumulative Review 110. You observe that sugar dissolves more quickly in hot tea

than in iced tea. You state that higher temperatures increase the rate at which sugar dissolves in water. Is this statement a hypothesis or a theory? Why? (Chapter 1)

111. Write the electron configuration for each of the follow-ing atoms. (Chapter 5)

a. fluorine c. titanium b. aluminum d. radon 112. Explain why the gaseous nonmetals exist as diatomic

molecules, but other gaseous elements exist as single atoms. (Chapter 8)

113. Write a balanced equation for the reaction of potassium with oxygen. (Chapter 9)

114. What is the molecular mass of U F 6 ? What is the molar mass of U F 6 ? (Chapter 10)

Percent Composition ofSome Organic Compounds

10

20

30

40

50

Perc

ent

by m

ass

0Ethanol Acetaldehyde Butanoic acidFormaldehyde

Compound name

52.2

13.0

34.840.0

6.7

53.3 54.5

9.1

36.4

54.5

9.1

36.4

%C%H%O

■ Figure 11.16

115. Figure 11.16 gives percent composition data for several organic compounds. (Chapter 10)

a. How are the molecular and empirical formulas of acetaldehyde and butanoic acid related?

b. What is the empirical formula of butanoic acid?

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Standardized Test Practice glencoe.com

Standardized Test PracticeMultiple Choice 1. D 2. A 3. C 4. B 5. C 6. D 7. C 8. B

Standardized Test PracticeCumulative

Multiple Choice

1. Stoichiometry is based on the law ofA. constant mole ratios.B. Avogadro’s constant.C. conservation of energy.D. conservation of mass.

Use the graph below to answer Questions 2 to 5.Supply of Various Chemicalsin Dr. Raitano’s Laboratory

Na2CO3500.0 gNaCl

700.0 g

Ca(OH)2300.0 g

NaH2PO4350.0 g

KClO3200.0 gAgNO3

100.0 g

2. Pure silver metal can be made using the reaction shown below.

Cu(s) + 2AgN O 3 (aq) → 2Ag(s) + Cu(N O 3 ) 2 (aq)How many grams of copper metal will be needed to use up all of the AgN O 3 in Dr. Raitano’s laboratory?A. 18.70 g C. 74.7 gB. 37.3 g D. 100 g

3. The LeBlanc process is the traditional method of manufacturing sodium hydroxide. The equation for this process is as follows.

N a 2 C O 3 (aq) + Ca(OH ) 2 (aq) → 2NaOH(aq) + CaC O 3 (s)Using the amounts of chemicals available in Dr. Raitano’s lab, what is the maximum number of moles of NaOH that can be produced?A. 4.05 mol C. 8.097 molB. 4.72 mol D. 9.43 mol

4. Pure O 2 gas can be generated from the decomposi-tion of potassium chlorate (KCl O 3 ):

2KCl O 3 (s) → 2KCl(s) + 3 O 2 (g)

If half of the KCl O 3 in the lab is used and 12.8 g of oxygen gas is produced, what is the percent yield of this reaction?A. 12.8% C. 65.6%B. 32.7% D. 98.0%

5. Sodium dihydrogen pyrophosphate (N a 2 H 2 P 2 O 7 ), more commonly known as baking powder, is manufactured by heating Na H 2 P O 4 to a high temperature.

2Na H 2 P O 4 (s) → N a 2 H 2 P 2 O 7 (s) + H 2 O(g)

If 444.0 g of N a 2 H 2 P 2 O 7 is needed, how much more Na H 2 P O 4 will Dr. Raitano have to buy to make enough N a 2 H 2 P 2 O 7 ?A. 0.00 gB. 94.0 gC. 130.0 gD. 480 g

6. Red mercury(II) oxide decomposes at high temperatures to form mercury metal and oxygen gas.

2HgO(s) → 2Hg(l) + O 2 (g) If 3.55 mol of HgO decomposes to form 1.54 mol

of O 2 and 618 g of Hg, what is the percent yield of this reaction?A. 13.2%B. 42.5%C. 56.6%D. 86.8%

Use the diagram below to answer Questions 7 and 8.

Y

Y YY

YYY

Y

Y

Y ZZZ

Z

ZZZ

Z

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

Z WWW

WW

WWW

WW

WWW

WW

WWW

WW

WWW

WW

WWW

WWY

ZZ

Z

YY

Y

X X X X X X X X X X X X X XX X X X X X X X X X X X X X

1

PERIODIC TABLE

2

3 4 5 6 7 8 9 10 11 12

13 14 15 16 17

18

7. Which elements tend to have the largest atomic radius in their periods?A. W C. YB. X D. Z

8. Elements labeled W have their valence electrons in which sublevel?A. s C. dB. p D. f

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Standardized Test Practice glencoe.com

Short Answer 9. 27 moles N 2

Extended Response 10. Data should form an approximately lin-

ear relationship with a few jagged edges, similar to Figure 6.16 on page 191. .

11. Ionization generally increases as you move across a period or row in the peri-odic table. Elements in the first few fami-lies have only 1 or 2 valence electrons, which are relatively easy to remove because this will result in a complete outer shell. Elements on the right side of the periodic table have very high ioniza-tion energies because their outer shells are nearly filled, therefore making it more likely for these elements to gain a few electrons rather than lose many.

SAT Subject Test: Chemistry 12. B 13. A 14. C 15. B 16. D 17. B

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Short Answer SAT Subject Test: Chemistry

9. Dimethyl hydrazine (C H 3 ) 2 N 2 H 2 ignites on contact with dinitrogen tetroxide ( N 2 O 4 ).

(C H 3 ) 2 N 2 H 2 (l) + 2 N 2 O 4 (l) → 3 N 2 (g) + 4 H 2 O(g) + 2C O 2 (g)

Because this reaction produces an enormous amount of energy from a small amount of reactants, it was used to drive the rockets on the Lunar Excursion Modules (LEMs) of the Apollo space program. If 18.0 mol of dinitrogen tetroxide is con-sumed in this reaction, how many moles of nitrogen gas will be released?

Extended Response

Use the table below to answer Questions 10 and 11.

First Ionization Energy of Period 3 Elements

Element Atomic Number 1st Ionization Energy, kJ/mol

Sodium 11 496

Magnesium 12 736

Aluminum 13 578

Silicon 14 787

Phosphorus 15 1012

Selenium 16 1000

Chlorine 17 1251

Argon 18 1521

10. Plot the data from this data table. Place atomic numbers on the x-axis.

11. Summarize the general trend in ionization energy. How does ionization energy relate to the number of valence electrons in an element?

12. How much cobalt(III) titanate (C O 2 Ti O 4 ), in moles, is in 7.13 g of the compound?A. 2.39 × 1 0 1 molB. 3.10 × 1 0 -2 molC. 3.22 × 1 0 1 molD. 4.17 × 1 0 -2 molE. 2.28 × 1 0 -2 mol

Use the pictures below to answer Questions 13 to 17.

A. D.

B. E.

C.

13. Hydrogen sulfide displays this molecular shape.

14. Molecules with this shape have four shared pairs of electrons and no lone pairs of electrons.

15. This molecular shape is known as trigonal planar.

16. Carbon dioxide displays this molecular shape.

17. This molecular shape undergoes s p 2 hybridization.

NEED EXTRA HELP?

If You Missed Question . . . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Review Section . . . 11.1 11.2 11.2 11.4 11.3 11.4 6.3 5.3 11.2 6.3 6.3 10.3 8.4 8.4 8.4 8.4 8.4

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