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StoichiometricCalculations

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Stoichiometry Stoichiometry is the quantitative determination of reactants or products using a balanced chemical equation.

We can interpret a balanced equation in several ways.

The most common way is to interpret it as indicating the number of moles of each substance. For instance, this equation,

3 H2 + N2 ® 2 NH3

can be read as:3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to yield 2 moles of ammonia, or

3 mol H2 + 1 mol N2 to yield 2 mol NH3

The coefficients in the balanced equation give the ratio of moles of reactants and products. Therefore, a balanced chemical equation is needed to perform any stoichiometric calculation.

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Since the coefficients in the balanced equation represent moles of substances, then we can write "mole ratios" which show relative amounts of reactants and products.

Here is an example showing how to write "mole ratios" :

Consider the reaction that produces ammonia, NH3.

N2 + 3 H2 ® 2 NH3

Here are three different mole ratios based on this equation.

1 mol N2

3 mol H2

1 mol N23 mol H2

2 mol NH3 2 mol NH3

Stoichiometric Calculations with Moles

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1 mol N2

3 mol H2

1 mol N23 mol H2

2 mol NH3 2 mol NH3


The above ratios can be used to determine the quantity of any reactant or product.

N2 + 3 H2 ® 2 NH3

For every 1 mol of N2,· you would need 3 mol of H2 to completely react with it, and· you would produce 2 mol of NH3

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N2 + 3 H2 ® 2 NH3

In fact, there are three more mole ratios that can be written simply by taking the reciprocal of the first three.

1 mol N2

3 mol H2

1 mol N23 mol H2

2 mol NH3 2 mol NH3

1 mol N2

3 mol H2

3 mol H2 1 mol N2

2 mol NH3 2 mol NH3


It is not important to write down all of the possible mole ratios for a given reaction. But you do need to know how to create a mole ratio for a specific problem.

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N2 + 3 H2 ® 2 NH3

Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?

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Stoichiometric Calculations with Moles Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?

Sample Problem #1 - Solution· In the problem, circle the "given" quantity and underline the "wanted" quantity.

· Write the "given" quantity. 4 moles of N2 · Select the mole ratio that has the "given" unit in the denominator.

Choose the mole ratio that has the "wanted" unit in the numerator.

1 mol N2

3 mol H2

1 mol N2

2 mol NH3

1 mol N2

3 mol H2

1 mol N2

2 mol NH3

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Stoichiometric Calculations with Moles Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?

Sample Problem #1 - Solution (con't)

We are now ready to set up our problem: 4 moles of N2

Just as you multiply fractions, multiply the numbers across the top and divide by any numbers in the denominators.

1 mol N2

2 mol NH3x

1

1 mol N2

2 mol NH3x

1

4 moles of N2= 8 mol NH3

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N2 + 3 H2 ® 2 NH3

Sample Problem #2How many moles of H2 will be needed to react with 15 mol N2?

Stoichiometry Calculations with Moles

The same approach can tell you how many moles of one reactant is needed to completely react with another reactant.

1 mol N2

3 mol H2

1 mol N23 mol H2

2 mol NH3 2 mol NH3

1 mol N2

3 mol H2

3 mol H2 1 mol N2

2 mol NH3 2 mol NH3

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1 mol N2

3 mol H2

1 mol N23 mol H2

2 mol NH3 2 mol NH3

1 mol N2

3 mol H2

3 mol H2 1 mol N2

2 mol NH3 2 mol NH3


Sample Problem #2- Solution

How many moles of H2 will be needed to react with 15 mol N2?

15 moles of N2 3 mol H2

1 mol N21x = 45 mol H2

We choose the mole ratio that has mol N2 in the denominator and mol H2 in the numerator.

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= 4 mol Al2O3

= 6 mol O2 x 2 molAl2O3 3 mol O2

1 What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

Answer

A 2 mol

B 3 mol

C 4 mol

D 18 mol

E 24 mol

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= 18 mol O2

2 How many moles of O2 would be required to create 12 moles of Al2O3?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

= 12 mol Al2O3 x 3 mol O2 2 mol Al2O3

Answer

A 3 mol

B 4 mol

C 8 mol

D 18 mol

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8mol Al x 3 mol O2 ----------- 4 mol Al

= 6 mol O2

3 How many moles of O2 would be required to completely react with 8 moles of Al?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

Answer

A 2 mol

B 3 mol

C 6 mol

D 12 mol

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4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol Fe?

4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)

2.4 mol Fe x 3 mol O2 ----------- 4 mol Fe

= 1.8 mol O2

Answer

A 1.2 mol

B 1.8 mol

C 2.4 mol

D 3.2 mol

E 4.8 mol

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5 How many moles of Al are needed to react completely with 1.2 mol FeO?

1.2 mol FeO x 2 mol Al ----------- 3 mol FeO

= 0.8 mol Al

Answer

2 Al (s) + 3 FeO (g) ® 3 Fe (s) + Al2O3 (s)

A 0.8 mol

B 1.2 mol

C 1.6 mol

D 2.4 mol

E 4.8 mol

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6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride?

CaCl2 (s) ® Ca (s) + Cl2 (g)

8 mol CaCl2 x 1 mol Ca ---------------- 1 mol CaCl2

= 8 mol Ca

Answer

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8 mol Ca8 mol Cl2Total of 16 moles

7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of cal cium chloride?

CaCl2 (s) ® Ca (s) + Cl2 (g)

Answer

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Silver and nitric acid react according to the following balanced equation:

3 Ag(s) + 4 HNO3(aq) à 3 AgNO3(aq) + 2 H2O(l) + NO(g)

A. How many moles of silver are needed to react with 40 moles of nitric acid, HNO3?

B. Using 40 mol HNO3, how many moles of silver nitrate will be produced?

C. Using 40 mol HNO3, how many moles of water will be produced?

D. Using 40 mol HNO3, how many moles of nitrogen monoxide will be made?


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Answers:

A) 30 mol Ag

B) 30 mol AgNO3

C) 20 mol H2O

D) 10 mol NO

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2 N2H4(l) + N2O4(l) à 3 N2(g) + 4 H2O(g)

A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen?

B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen?

C. How many moles of water are produced when 57 moles of nitrogen are made?


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Answers:

A) 38 mol N2H4

B) 19 mol N2O4

C) 76 mol H2O

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Stoichiometry

So far, we have interpreted the coefficients in a balanced equation as indicating the number of moles of each substance. However, we said before that there are many different ways to interpret a balanced equation.

The coefficients in a balanced equation can represent· moles· representative particles (atoms, molecules, formula units)· liters of gas (at constant temperature & pressure)

The chart on the next slide shows these relationships.

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3 H2 N2 2 NH3

3 moles of H2 1 mole of N2 2 moles of NH3

3 molecules of hydrogen (H2)

1 molecule of nitrogen (N2)

2 molecules of ammonia (NH3)

3 L of H2 * 1 L of N2 * 2 L of NH3 *

* at same Temperature & Pressure

Interpreting a Balanced Equation

3H2 + N2 à 2NH3

· moles· representative particles (atoms, molecules, formula units)· liters of gas *

The coefficients in a balanced equation can represent

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3H2 + N2 à 2NH3

· In terms of moles, the equation above showsa total of 4 moles reacting to yield 2 moles of product.

· In terms of particles, a total of 4 molecules react to form 2 molecues.

· In terms of liters, a total of 4 liters of gas react to form 2 liters of gas (at STP, 89.2 L of gas react to yield 44.8 L).

The number of moles, particles and liters before the arrow is not the same as after the arrow. Therefore, these quantities are not conserved in a balanced equation.

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3H2 + N2 à 2NH3

We have seen various ways to interpret a balanced equation.

Recall that a balanced equation is one in which both sides have the same number of each kind of atom. As a result, the Law of Conservation of Mass is obeyed, since no atoms can be created nor destroyed.

Therefore, the quantities that are conserved in a balanced equation are mass and atoms. Conserved means that the amount that appears before the arrow is the same as the amount after the arrow.

Only mass and the number of atoms are conserved in a balanced equation.

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3 H2 N2 2 NH3

6 atoms of hydrogen (H)

2 atoms of nitrogen (N)

6 atoms of hydrogen (H)and

2 atoms of nitrogen (N)

3 mol x 2 g/mol

= 6 g of H2

1 mol x 28 g/mol

= 28 g of N2

2 mol x 17 g/mol

= 34 g of NH3


3H2 + N2 à 2NH3

Only mass and the number of atoms are conserved in a balanced equation.

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8 In a chemical reaction that is represented by a balanced equation, the quantities that are conserved are

A mass and molecules

B mass and atoms

C moles and liters

D moles and molecules

E mass and liters

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9 In a chemical reaction that is represented by a balanced equation, the quantities that are NOT conserved are

A mass and moles

B mass and atoms

C moles and particles

D mass and atoms

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Stoichiometry Calculations with ParticlesWe will now consider a second interpretation of a balanced equation in terms of particles (which could be molecules, atoms, or formula units).

Multiplying the number of moles by 6.02 x 1023 yields the numbers of particles. So a formula that's balanced for moles must also be balanced for particles.

3 H2 + N2 ® 2 NH3can be read as:

3 molecules of H2 plus 1 molecule of N2 yields 2 molecules of NH3

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Stoichiometry Calculations with Particles

Solving stoichiometry problems with particles is similar to those with moles. One big difference is that your answers for particles cannot be fractions or decimal numbers; they must be whole numbers.

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Stoichiometry Calculations with ParticlesSo here are the six ratios that you could use for solving problems involving particles:

3 H2 + N2 ® 2 NH3

2 molecules NH3

3 molecules H2

3 molecules H2

3 molecules H2

1 molecule N2

1 molecule N2

1 molecule N2

2 molecules NH3

2 molecules NH3

and their reciprocals

3 molecules H2

2 molecules NH3

1 molecule N2

Note that there is no abbreviation for "molecule."

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Stoichiometry Calculations with Particles

Sample Problem #3How many molecules of N2 are needed to completely react with 24 molecules of H2?

Sample Problem #3 - Solution · In the problem, circle the "given" quantity and underline the "wanted" quantity.

How many molecules of N2 are needed to completely react with 24 molecules of H2?

· Write the "given" quantity.

24 molecules of H2 1

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Stoichiometry Calculations with ParticlesSample Problem #3How many molecules of N2 are needed to completely react with 24 moles of H2?

Sample Problem #3 - Solution (con't)· Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.

2 molecules NH3

3 molecules H2

3 molecules H2

3 molecules H2

1 molecule N2

1 molecule N2

1 molecule N2

2 molecules NH3

2 molecules NH3

3 molecules H2

2 molecules NH3

1 molecule N2

24 molecules of H2 1

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Stoichiometry Calculations with ParticlesSample Problem #3How many molecules of N2 are needed to completely react with 24 moles of H2?

Sample Problem #3 - Solution (con't)· Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.

24 molecules of H2 1 3 molecules H2

1 molecule N2x = 8 molecules N2

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10 What is the largest number of of Li3N formula units that could result from reacting 6 N2 molecules?

6 Li (s) + N2 (g) ® 2 Li3N (s)

6 molecules N2 x 2 formula units Li3N ---------------------- 1 molecule N2

= 12 formula units Li3N

Answer

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11 How many N2 molecules would be required to create 4 Li3N formula units?

6 Li (s) + N2 (g) ® 2 Li3N (s)

4 formula units Li3N x 1 molecule N2 ---------------------- 2 formula units Li3N

= 2 molecules N2

Answer

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12 How many Li atoms would be required to completely react with 3 N2 molecules?

6 Li (s) + N2 (g) ® 2 Li3N (s)

3 molecules N2 x 6 atoms Li ---------------------- 1 molecule N2

= 18 atoms Li

Answer

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Stoichiometry Calculations with Volume

So far, we have seen that coefficients in a balanced equation can represent moles or representative particles. A third interpretation applies only to gases involved. If all the gases are at the same temperature and pressure, then the coefficients give the relative volumes of gas in liters.

CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)

This formula can be read as:*

1 L of CS2 + 3 L of O2 yields 1 L of CO2 + 2 L of SO2

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Stoichiometry Calculations with VolumeUsing coefficients to represent volume only works when comparing two gases in a formula. It cannot be used to compare a gas to a liquid or solid.

CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)

At STP conditions:

1 x 22.4 L + 3 x 22.4 L ® 1 x 22.4 L + 2 x 22.4 L

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Using this interpretation we can directly answer questions about the relative volumes of gas reactants and products.

For the next problem, we will not write out all of the ratios ahead of time.

Sample Problem #4How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below?

CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)

Stoichiometry Calculations with Gases

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Sample Problem #4 - Solution

How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below?

CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)

Stoichiometry Calculations with Gases

· Write the "given" quantity.· Create a ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.

6 liters of O2

3 L of O2

2 L of SO2x = 4 L of SO2

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13 The coefficients in a balanced equation can represent

A mass, in grams

B volume, in liters (gases only)

C representative particles

D Both A and B

E Both B and C

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= 8 L of H2O

14 How many liters of H2O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)?

2 H2 (g) + O2 (g) ® 2 H2O (g)

Answer

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15 The equation below shows the decomposition of solid lead (II) nitrate, Pb(NO3)2. How many liters of O2 will be produced when 12 L of NO2 are formed?

2 Pb(NO3)2 (s) ® 2 PbO (s) + 4 NO2 (g) + O2 (g)

= 3 L of O2

= 12 L NO2 x 1 L O2 4 L NO2

Answer

A 1.0 L

B 2.0 L

C 3.0 L

D 4.0 L

E 12 L

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= 250 L of O2

16 What volume of methane is needed to completely react with 500 L of O2 at STP? You must first balance the equation.

__ CH4 + ___ O2 --> ___ CO2 + ___ H2O

= 500L O2 x 1 L CH4 2 L O2

Answer

A 125 L

B 250 L

C 500 L

D 1000 L

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= 36 L O2 x 4 L NO2 7 L O2

= 21 L of NO2

17 How many liters of NO2 (g) will be created from reacting 36 L of O2 (g) with a sufficient amount of NH3 (g)?

4 NH3 (g) + 7 O2 (g) ® 4 NO2 (g) + 6 H2O (g)

Answer

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So far, we have seen stoichiometry problems that can be solved in one step, using only coefficients. These are: · mole-mole problems, · particle-particle problems and · volume-volume problems.

Coefficients can represent moles, particles, or liters.

When mass is involved, however, we must now take into consideration the fact that all particles have a different molar mass.

Coefficients do not represent mass in grams.

Stoichiometric Calculations with Mass

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3 H2 N2 2 NH3

3 moles of H2 1 mole of N2 2 moles of NH3

3 mol x 2 g/mol

= 6g of H2

1 mol x 28 g/mol

= 28g of N2

2 mol x 17 g/mol

= 34g of NH3

3 H2 + N2 ® 2 NH3

Using this example once again, it is clear that 3 grams of H2 + 1 gram of N2 will not equal 2 grams of NH3.

Coefficients do not represent mass in grams.

6 g of H2 + 28 g of N2 will equal 34 g of NH3

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Consider this question asked earlier:What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2?

4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)

Now, instead of asking for moles of Al2O3 , consider this question:

Sample Problem #5What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2?

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What is the mass, in grams, of Al2O3 that could result from reacting

6 moles of O2?

We proceed similarly as before, changing from moles of the "given" quantity to the moles of the "wanted" quantity.


4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)

6 moles of O2

3 mol O2

2 mol Al2O3x1

= 4 mol Al2O3

This answer becomes the new "given" which we simply convert to grams, using the molar mass of Al2O3.

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What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2?


4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)

4 mol Al2O3

1 mol Al2O3

102 g Al2O3x1

= 408 g Al2O3

Remember: Do not use coefficients when converting between moles and mass. The molar mass is for only ONE mole!

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An alternative way to solve the previous problem is to set-up both ratios, then perform the calculation without stopping after the first step.

6 moles of O2

3 mol O2

2 mol Al2O3x1 1 mol Al2O3

102 g Al2O3x = 408 g Al2O3

The key is to make sure that the unit in any numerator appears in the next denominator so that units will cancel out.

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= 18 mol O2 x 32 g O2 1 mol O2

= 576 g O2

18 What mass, in grams, of O2 would be required to create 12 moles of Al2O3?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

= 12 mol Al2O3 x 3 mol O2 2 mol Al2O3

Answer

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= 192 g O2

8mol Al x 3 mol O2 ----------- 4 mol Al

= 6 mol O2 x 32 g O2 1 mol O2

Answer

19 How many grams of O2 would be required to completely react with 8 moles of Al?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

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20 When iron rusts in air, iron (III) oxide is produced. How many grams of oxygen react with 2.4 mol Fe?

4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)

2.4 mol Fe x 3 mol O2 ----------- 4 mol Fe

= 1.8 mol O2 x 32 g O2 1 mol O2

= 57.6 g O2

Answer

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21 How many grams of Al are needed to react completely with 1.2 mol FeO?

1.2 mol FeO x 2 mol Al ----------- 3 mol FeO

= 0.8 mol Al x 27 g Al 1 mol Al

= 21.6 g Al

Answer

2 Al (s) + 3 FeO (g) ® 3 Fe (s) + Al2O3 (s)

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Mass-Mass Calculations

Given Find

We started this unit by using coefficients in a balanced equation to solve mole-mole problems.

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Next, we extended these problems by calculating the mass of the "wanted" quantity.


Given

Find

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Now, we will solve problems when both the "given" quantity and the "wanted" quantity are in grams. Recall that you cannot simply use the coefficients to relate mass in grams.


Grams of substance A Grams of

substance B

Moles of substance A

Moles of substance B

Use molar mass of A Use molar

mass of B

Use coefficients of A and B

from balanced equation

GivenFind

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A typical "mass-mass" problem is has three steps as outlined below. Remember that coefficients are used ONLY between moles.



substance B




mass of B



GivenFind

Step 1

Step 2

Step 3

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Sample Problem #6Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.

N2 + 3H2 ---> 2NH3

Sample Problem #6 - SolutionCalculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.

We will proceed using these steps:

a) convert 5.4 g H2 to moles H2

b) convert moles H2 to moles NH3

c) convert moles NH3 to grams NH3

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a) convert 5.4 g H2 to moles H2

c) convert moles NH3 to grams NH3

b) convert moles H2 to moles NH3


gH2 --> mol H2 --> mol NH3 --> g NH3

5.40 g H2 1 mol H22.0 g H2

17.0 g NH31 mol NH3

2 mol NH33 mol H2

X = 31 g NH3X X


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Sample Problem #7How many grams of water can be produced from the combustion of 1.00 g of glucose?

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O


Our general strategy isStep (a): convert mass of glucose to moles of glucose· use the molar mass of glucose

Step (b): convert moles glucose to moles water· use the ratio of coefficients in the balanced equation

Step (c): convert moles of water to grams of water· use the molar mass of water


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C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O


1mol glucose180.0 g glucose

5.56 x 10-3 mol glucose 6 mol water

1 mol glucose3.33 x 10-2 mol water

18.0 g water

1 mol water

0.600 g water1.00 g glucose

X

X

X

(a) (c)

no direct calculation

(b) for every 1 mol glucoseyou get 6 mols of water

Sample Problem #7 - Solution (con't)How many grams of water can be produced from the combustion of 1.00 g of glucose?

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1.00 g glucose

11mol glucose

180.0 g glucosex x 6 mol water

1 mol glucose

18.0 g water

1 mol waterx


substance B




mass of B



GivenFind

= 0.600 g water

Again, the key to this method is making sure that the unit in any numerator appears in the next

denominator so that units will cancel out.

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22 What mass of Na is produced by the decomposition of 6.5 g NaN3?

A 2.3 g

B 4.6 g

C 6.5 g

D 23 g

E 46 g

2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)

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23 How many grams of Al2O3 will be created from reacting 54 g of Al with a sufficient amount of O2?

4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)

A 27 g

B 54 g

C 102 g

D 204 g

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24 How many grams of Mg must react in order to to create 160 g of MgO?

2 Mg (s) + O2 (g) ® 2 MgO (s)

A 48 g

B 80 g

C 96 g

D 160 g

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25 Approximately how many grams of oxygen are needed to react with 240 g of Mg?

2 Mg (s) + O2 (g) ® 2 MgO (s)

A 80 g

B 120 g

C 160 g

D 240 g

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Mixed Stoichiometry CalculationsThus far, our problem-solving has focused on one of four main types. Most practical applications of chemistry are not this narrowly defined. Many problems in advanced chemistry give a quantity in one unit (e.g. moles, grams, or liters) and ask for the proportional quantity in a different unit.

For example, this problem gives a quantity in moles, but asks for mass, in grams.

How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?

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Mixed Stoichiometry CalculationsEvery type of stoichiometry calculation may be solved by following this map.

(1) From left to right, we convert any "Given" substance to moles. (2)Next, using the mole ratio created with coefficients, one can calculate

the moles of the "Wanted" quantity. (3) Finally, if necessary, moles can be converted to either

particles, mass or volume.(1)

(2)

(3)1 mol G

6.02 x10-23

represenative particles of G X

1 mol G

mass G

mass of G X

1 mol G

22.4 L Gvolume of Gat STP

X

mass W

1 mol W

massof W=

22.4 L W

1 mol WVolume of W at STP=

b mol W

a mol Gmol G X mol W

x

x

6.02 x 1023

1 mol W

representative particles of W=x

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N2 + 3H2 ---> 2NH3

Sample Problem #8How many grams of ammonia can be produced by using10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?


Since the "Given" quantity is already in moles, we can skip to step (2).

Overview:mol N2 ---> mol NH3 ---> grams NH3

Mixed Stoichiometry Calculations

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10 mol N2 2 mol NH3

1 mol N2

x = 340 g NH3x1 mol NH3

17 g NH3

N2 + 3H2 ---> 2NH3



Remember: Do not use coefficients when using molar mass.

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Sample Problem #9The airbag in a car generates a relatively large amount of gas quickly through the following reaction.

2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)

How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)?


How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)?


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1 mol G

6.02 x10-23

represenative particles of GX

1 mol G

mass G

mass of G X

1 mol G

22.4 L Gvolume of Gat STP

X

mass W

1 mol W

massof W=

22.4 L W

1 mol WVolume of W at STP=

b mol W

a mol Gmol G X mol W

x

x

6.02 x 1023

1 mol W

representative particles of W=x

(1)

(2) (3)

2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)

Sample Problem #9 - Solution OverviewStep (1) - Liters of N2 --> Moles of N2Step (2) - Moles of N2 --> Moles of NaN3Step (3) - Moles of NaN3 --> Grams of NaN3


Remember that coefficients are used ONLY between moles.

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2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)

Sample Problem #9 - Solution Step (1) - Liters of N2 --> Moles of N2


Step (2) - Moles of N2 --> Moles of NaN3

Step (3) - Moles of NaN3 --> Grams of NaN3

36 L N2 1 mol N2

22.4 L N2x = 1.6 mol N2

1

2 mol NaN3

3 mol N2x

1.6 mol N2

1 = 1.1 mol NaN3

65 g NaN3

1 mol NaN3x

1.1 mol NaN3

1 = 72 g NaN3

Remember that coefficients are used ONLY between moles.

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26 How many liters of O2 (g) at STP are required to create 102 g of Al2O3 (s)?

4 Al (s) + 3O2 (g) ® 2 Al2O3 (s)

x x2 mol Al2O3

3 mol O2 102 g Al2O3 1 mol Al2O3 102 g Al2O3 1 mol O2

22.4 L O2 x1

= 33.6 L O2

Answer

A 11.2 L

B 22.4 L

C 33.6 L

D 44.8 L

E 67.2 L

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27 How many grams of Cl2 are needed to react with 1 mol of antimony, Sb?

2 Sb + 3 Cl2 à 2 SbCl3

A 71 g

B 107 g

C 142 g

D 213 g

x x1 mol Sb 3 mol Cl2 2 mol Sb 1 mol Cl2

71 g Cl2

1= 107 g Cl2

Answer

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A. How many manganese atoms are produced if 55 moles of MnO2 react with excess aluminum.

B. How many moles of aluminum oxide are made if 3580 g of manganese oxide are consumed?

3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)

*Mixed Stoichiometry Practice

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3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)

*Mixed Stoichiometry Practice

C. How many moles of manganese oxide will react with 5.33 x 1025 atoms of aluminum?

D. If 4.37 moles of aluminum are consumed, how many formula units of aluminum oxide are produced?

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Answers:

A) 3.3 x 1025 atoms Mn

B) 27.5 mol Al2O3

C) 66.4 mol MnO2

D) 1.32 x 1024 formula units Al2O3

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The compound tristearin (C57H110O6) is a type of fat which camels store in their hump. Besides being a source of energy, the fat is a source of water for the camel because when the fat is burned, the following combustion reaction occurs:

2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)

A. At STP, what volume of O2 is required to consume 0.64 moles of tristearin?

B. At STP, what volume of carbon dioxide is produced in Part A?

**Mixed Stoichiometry Practice

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2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)

C. If 22.4 L of oxygen is consumed at STP, how many moles of water are produced?

D. Find the mass of tristearin required to produce 55.56 moles of water.

**Mixed Stoichiometry Practice

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Answers:

A) 1168 L O2 (or 1200 L)

B) 817 L CO2 (or 820 L)

C) 0.00414 mol H2O

D) 899 g C57H110O6

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Limiting Reagent & Excess Reagent

Percent Yield & Theoretical Yield

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How Many Cookies Can I Make?

· You can make cookies until you run out of one of the ingredients.· Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

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In this example the egg would be the limiting reactant, because it will limit the amount of cookies you can make.


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Recipe for Butter Cream Cookies

1 c. butter 1 c. sugar 2 T. vanilla

1-3oz. pkg cream cheese

1 egg yolk 2-1/2 c. sifted cake flour

This recipe will make 5 dozen cookies.

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28 How many dozen cookies could you make with 3 c. butter and enough of all the other ingredients?

A 1 doz

B 3 doz

C 5 doz

D 10 doz

E 15 doz





Yields 5 dozen cookies.

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29 How many dozen cookies could you make with 10 c. sifted cake flour and enough of all the other ingredients?

A 2 doz

B 5 doz

C 10 doz

D 20 doz

E 40 doz






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30 How many dozen cookies could you make with 1 T. vanilla and enough of all the other ingredients?

A 1 doz

B 2 doz

C 2-1/2 doz

D 5 doz

E 10 doz






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Limiting Reagents & Theoretical Yield

· The limiting reactant, or limiting reagent, is the reactant present in the smallest stoichiometric amount.

· This is not necessarily the one with the smallest mass.

· The limiting reactant is the reactant you’ll run out of first, and it is the one that determines the maximum amount of product that can be made.

· Theoretical yield is the the maximum amount of product that can be made, based on the limiting reagent.

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Limiting Reagents

In the three previous Senteo questions, the amount given in the problem was the limiting reagent.

· 3 c butter· 10 c. sifted cake flour· 1T. vanilla

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Limiting Reagents

Limiting reagent Excess reagent Theoretical yield

3 c. butter all other ingredients 15 doz.

10 c. sifted cake flour

all other ingredients 20 doz.

1 T. vanilla all other ingredients 2-1/2 doz.

Note that in every case, the theoretical yield is determined by the limiting reagent, not the excess reagents.

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31 When two substances react, the one that is used up first is called the:

A determining reagent

B limiting reagent

C unlimited reagent

D excess reagent

E reactive reagent

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Limiting reagent problems are different from those done previously in that two quantities are given, instead of just one.

It is your job to figure out which reactant is limiting because that will determine the maximum yield.

There are a variety of methods one can use to determine which reactant is the limiting one. Once you have identified the limiting reagent, then the other reactant is the excess reagent.

Limiting Reagents

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Definitions· Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction

· Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction

· Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent

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This is how you identify which is the limiting reagent:

1. Write a balanced equation.

2. Use the given amount of each reactant to calculate the amount of product that could be formed. Use the problem-solving methods for stoichiometry that you learned earlier in this unit.

3. Compare the two amounts of product that could be made. The smaller of the two amounts indicates the maximum amount of product that could be made and is called theoretical yield. The larger of the two amounts is irrelevant and meaningless.

4. Whichever reactant yields the smaller amount of product is thus the limiting reagent.


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Sample Problem #10Consider the reaction between hydrogen and oxygen to yield water. 2 H2 (l) + O2 (g) --> 2 H2O (l)

Starting with 10 molecules of H2 and 7 molecules of O2, which reactant will run out first?

Limiting Reagent & Excess ReagentO2 H2

Sample Problem #10 - SolutionUse stoichiometry methods to calculate the following:

10 molecules of H2 will yield ___ molecules of H2O

7 molecules of O2 will yield ___ molecules of H2O

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10 molecules H2

10 molecules of H2 will yield __10_ molecules of H2O

7 molecules O2

7 molecules of O2 will yield __14_ molecules of H2O

Limiting Reagent & Excess Reagent2 H2 (l) + O2 (g) --> 2 H2O (l)

2 molecules H2 x

x

2 molecules H2O

2 molecules H2O

1 molecule O2

Now, we compare the two amounts.

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10 molecules of H2 will yield __10_ molecules of H2O

7 molecules of O2 will yield __14_ molecules of H2O

· The smaller amount is the theoretical yield: in reality, 10 molecules of water can be produced.

· 10 molecules of H2 is the limiting reagent

· 7 molecules of O2 is the excess reagent.

· The amount of 14 molecules of water cannot in fact be made; this would require 14 molecules of H2 which is not available. The amount of 14 molecules is meaningless and serves only to compare to the 10 molecules of water that can be produced.


2 H2 (l) + O2 (g) --> 2 H2O (l)

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Sample Problem #11What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen?

Sample Problem #11 - SolutionThis problem asks for theoretical yield; however, we must first determine which reagent is limiting.1. Write a balanced chemical equation. N2 + 3H2 ---> 2NH3

2. Calculate how much product can be made from the first "given" amount, 65 g nitrogen.

3. Calculate how much product can be made from the second "given" amount, 25 g hydrogen.


x x =65 g N2

28 g N2

1 mol N2

1 1 mol N2

2 mol NH3

1 mol NH3

17 g NH3x 79 g NH3

x x =25 g H2

2 g H2

1 mol H2

1 3 mol H2

2 mol NH3

1 mol NH3

17 g NH3x 142 g NH3

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Sample Problem #11What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen?

Sample Problem #11 - Solution (con't)4. Compare the two amounts. The smaller of the two amounts is the answer to the problem. This is called the theoretical yield.

Answer: 79 g of NH3 can be produced; nitrogen is the limiting reagent and hydrogen is the excess reagent.


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32 If you are provided with 75 molecules of H2 and 50 molecules of N2, what are the limiting reagent & theoretical yield for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)

A 75 molecules H2; 25 molecules NH3

B 75 molecules H2; 50 molecules NH3

C 50 molecules N2; 50 molecules NH3

D 50 molecules N2; 100 molecules NH3

E Not enough information is given.

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33 If you are provided with 50 molecules of H2 and 50 molecules of O2, which is the limiting reagent for the reaction:2 H2 (g) + O2 (g) --> 2 H2O (l)

A 50 molecules H2

B 50 molecules O2

C 50 molecules H2O

D There is no limiting reagent.


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34 If you are provided with 60 molecules of H2 and 20 molecules of N2, which is the limiting reagent for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)

A 60 molecules H2

B 20 molecules N2

C 50 molecules NH3



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35 If you are provided with 15 molecules of H2 and 10 molecules of N2, which is the limiting reagent for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)

A 15 molecules H2

B 10 molecules N2

C 20 molecules NH3



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36 Which of the following is NOT a true statement?

A The amount of product is determined by the limiting reagent.

B A balanced equation is necessary to determine which reagent is limiting.

C Some of the excess reagent is left over after the reaction is complete.

D The reactant that has the smallest given mass is the limiting reagent.

E Adding more of the limiting reagent will cause more product to be made.

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2Al(s) + 3Cl2(g) à 2AlCl327 g 71 g ? g

Sample Problem #12Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react.


1. See how much AlCl3 can be made from each reactant2. Smaller amount identifies the LR

Limiting Reagent & Theoretical Yield

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2Al(s) + 3Cl2(g) à 2AlCl327 g 71 g ? g


Sample Problem #12 - Solution (con't)27g Al will yield _______ g AlCl3

71g Cl2 will yield _______ g AlCl3

Therefore, the LR is _____________and the theoretical yield is ________ grams of AlCl3

27g Al will yield 134 g AlCl3

71g Cl2 will yield 89 g AlCl3

LR = 71 g Cl2TY = 89 g AlCl3

Answer

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2Al(s) + 3Cl2(g) à 2AlCl3100 g 100 g ? g

Sample Problem #13Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react.


1. See how much AlCl3 can be made from each reactant2. Smaller amount identifies the LR


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2Al(s) + 3Cl2(g) à 2AlCl3100 g 100 g ? g


Sample Problem #13 - Solution (con't)100g Al will yield ______ g AlCl3

100g Cl2 will yield _______ g AlCl3

Therefore, the LR is ____________and the theoretical yield of AlCl3 is ____________ grams.

100g Al will yield 494 g AlCl3

100g Cl2 will yield 125 g AlCl3

LR = 100 g Cl2TY = 125 g AlCl3

Answer

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You may sometimes be asked to calculate how much of the excess reagent is left over, after all of the limiting reagent has been consumed.

The general strategy is rather simple, once you have identified the excess reagent.

How to calculate excess reagent

It does not matter if the amounts are in moles or mass (grams), as long as you are consistent.

Excess Reagent

1) Write down how much of it you have at the start of the reaction.

2) Use stoichiometry (i.e. a balanced equation and the LR) to calculate how much of the excess reagent gets used up.

3) Subtract the answer from Step 2 from Step 1.

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BIG IDEA #1:

The limiting reagent determines the theoretical yield.

BIG IDEA #2:

The limiting reagent determines how much of the excess reagent gets used up.

Limiting Reagent, Excess Reagent & Theoretical Yield

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Consider this reaction which we have seen before:

2 H2 (l) + O2 (g) --> 2 H2O (l)

Sample Problem #14If we start with 10 molecules of H2 and 7 molecules of O2, how much of the excess reagent will be left over after the reaction is complete?

O2 H2

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Sample Problem #14 - SolutionWe had already determined (in Sample Problem #10) that hydrogen is the limiting reagent.

Therefore, oxygen is the excess reagent.and we must now complete the following:

O2 we start with: ________________O2 we use (based on LR): ________________O2 we have left over: __________________

2 H2 (l) + O2 (g) --> 2 H2O (l)O2 H2

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Sample Problem #14 - Solution (con't) O2 we start with: 7 molecules O2

O2 we use(based on LR*): 5 molecules O2

O2 we have left over: 2 molecules O2

2 H2 (l) + O2 (g) --> 2 H2O (l)

2 molecules H2 x 1 molecule O2

10 molecules H2 = 5 molecules O2 get used

1Always start with the LR here!

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Fe is the LR.O2 is the ER.You start with 12 molecules of O2 use up 9 of them, so 3 molecules are left over.


Try this one on your own:

4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)

Sample Problem #15If we start with 12 molecules of Fe and 12 molecules of O2, how much of the excess reagent will be left over after the reaction is complete?

Answer

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37 If you start with 16 mol H2 and 18 mol O2, how much of the excess reagent is left over?2 H2 + O2 ® 2 H2O

A 2 mol H2

B 7 mol H2

C 2 mol O2

D 10 mol O2

E Nothing is left over.

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38 If you start with 9 mol H2 and 5 mol N2, how much of the excess reagent is left over?

3 H2 + N2 ® 2 NH3

A 2 mol N2

B 4 mol N2

C 4 mol H2

D 6 mol H2


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39 If you start with 16 mol Fe and 12 mol O2, how much of the excess reagent is left over?

4 Fe + 3 O2 ® 2 Fe2O3

A 4 mol FeB 12 mol FeC 4 mol O2

D 9 mol O2


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____H2 + ____O2 ---> ____H2O

A. Calculate the theoretical yield (in grams) of water if 6 g H2 and 160 g O2 are combined in a reaction vessel.

B. How many grams of the excess reagent will be remaining in the vessel?

Limiting Reagent & Theoretical YieldPractice Problem 1

6 g H2 is the LR.160 g O2 is the ER.You start with (5 mol) 160 g of O2 and use up (1.5 mol) 48 g, so (3.5 mol) 112 g are left over.

Answer

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200 g AgNO3 is the LR.

7.10 x 1023 atoms Ag can be made.

According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate?

Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)

Limiting Reagent & Theoretical YieldPractice Problem 2 *

Answer

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At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? First, write a balanced equation.

Limiting Reagent & Theoretical YieldPractice Problem 3 *

50 g N2 is the LR.

40 L N2O can be made.

Answer

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___N2(g) + ___O2(g) à ___N2O5(g)

A. If you begin with 400 g of N2 and 800 g of O2, how many liters of N2O5 could be produced at STP?

B. Find the mass (in grams) of excess reagent left over at the conclusion of the reaction.


800 g O2 is the LR.400 g N2 is the ER.You start with (14.3 mol) 400 g of N2 and use up (10 mol) 280 g, so (4.3 mol) 120 g are left over.

Answer

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Carbon monoxide can be combined with hydrogen to produce methanol, CH3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152.5 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? (First, write a balanced equation.)


152.5 kg CO is the LR.

5446 mol = 174 kg CH3OHcan be made.

Answer

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Credit to Tom GreenboweChemical Education Group at Iowa State University

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Theoretical Yield & Actual Yield

We have recently seen that whenever you are given TWO amounts of reactants, you must first determine which is the limiting reagent.

BIG IDEA: The limiting reagent determines the

theoretical yield.

However, when experiments are carried out in the laboratory, the theoretical yield is not always easily obtained. The amount of product that is made when performing an experiment is called "Actual yield."

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Theoretical Yield vs. Actual Yield

Theoretical yield (TY) Actual yield (AY)

the maximum amount of product that can be made,

based on the stoichiometry (i.e. balanced equation) and

limiting reagent

the amount of product one actually produces and

measures in the laboratory; usually less than the TY

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· Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield).

· The efficiency of a reaction can be expressed using this ratio.

· For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%.

· Chemical manufacturers strive for high percent yields in their processes.

Actual YieldTheoretical Yield Percent Yield = x 100

Percent Yield

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Sample Problem #16Use the balanced equation to determine the percent yield if 5 g Mg reacts to actually form 8.1g of MgO

2 Mg (s) + O2 (g) à 2 MgO (s)


1) Since only one amount is given, then 5 g Mg must be the limiting reagent; oxygen is the excess reagent since there is no amount given.

2) Use 5 g to calculate the TY of MgO.

Percent Yield

5 g Mg24.3 g Mg

x 1 mol Mg

= 8.29 g MgO can be made1 2 mol Mg

x 2 mol MgO

1 mol MgO40.3 g MgO

x

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2 Mg(s) + O2(g) à 2 MgO(s)


3) Use percent yield formula

Actual YieldTheoretical Yield Percent Yield = x 100

8.1 g MgO8.3 g MgO Percent Yield = x 100

Percent Yield = 98%

Percent Yield

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40 What is the percent yield if the theoretical yield is 57.3 g and the actual yield is 50.9 g?

A 0.8883%

B 1.1257%C 88.83%D 112.57%

E 0.117%

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41 What is the actual yield if the theoretical yield is 16 g and the percent yield is 94.23%?

A 16.98 g

B 15.08 gC 100 gD 28%

E 1507.68

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Calcium hydroxide, Ca(OH)2, is also known as “slaked lime” and it is produced when water reacts with “quick lime,” CaO. If you start with 2.4 kg of quick lime, add excess water, and produce 2.06 kg of slaked lime, what is the percent yield of the reaction?

Stoichiometry Practice ProblemPractice Problem 6

· Is this a limiting reagent problem?· Is the 2.06 kg a theoretical yield or actual yield?· What quantity must you solve for?· Did you write a balanced equation?

TY = 3.17 kg Ca(OH)2

PY = (2.06/3.17) x 100= 65%

Answer

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TY = 533 g Fe

PY = (464 g Fe/533 g Fe) x 100= 65%

Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with aluminum to produce iron and aluminum oxide (Al2O3).

In one such reaction, 258 g of aluminum and excess rust produced 464 g of iron. What was the percent yield of the reaction?

Remember to first write a balanced chemical equation.


Answer

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TY = 5.6 x 104 L SO2

75% = (AY/5.6 x 104 L) x 100AY = (0.75)(5.6 x 104 L) AY = 4.2 x 104 L SO2

Use the balanced equation to find out how many liters of sulfur dioxide are actually produced at STP if 1.5 x 1027 formula units of zinc sulfide are allowed to react with excess oxygen and the percent yield is 75%.

2 ZnS(s) + 3 O2(g) à 2 ZnO(s) + 2 SO2(g)


Answer

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Definitions· Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction

· Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction

· Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent and a balanced chemical equation

· Actual yield - the amount of product that forms when a reaction is carried out in the laboratory

· Percent yield - the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percent; it is a measure of the efficiency of a reaction

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Outline:· General idea of stoichiometry· mole ratios· mole-mole problems· particle-particle problems· volume-volume· mole-mole-mass problems· mass-mass problems· mixed problems· LR/TY/PY· how to determine LR· using LR to obtain TY· calculate amount of ER left over· solve for PY or AY

stoichiometric calculations - njctlcontent.njctl.org/courses/science/chemistry/stoich... · 2...

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