chemical reactions chapter 4 reactants: zn + i 2 product: zni 2

CHEMICAL REACTIONSCHEMICAL REACTIONSCHAPTER 4CHAPTER 4

Reactants: Zn + I2 Product: ZnI2

2

Chapter 4 OutlineChapter 4 Outline• Chemical Equations

• Stoichiometry

–Limiting Reactants

–Chemical Analysis

3

Chemical EquationsChemical EquationsDepict the kind of Depict the kind of reactants reactants and and

products products and their relative amounts in a and their relative amounts in a

reactionreaction

4 Al(s) + 3 O4 Al(s) + 3 O22(g) 2Al(g) 2Al22OO33((s)s)

The numbers in the front are called The numbers in the front are called

stoichiometric coefficientsstoichiometric coefficients

The letters (s), (g), and (l) are the The letters (s), (g), and (l) are the

physical states of compounds.physical states of compounds.

4

Chemical EquationsChemical Equations

4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g) 2 Al2 Al22OO33(s)(s)

This equation meansThis equation means

4 Al atoms + 3 O4 Al atoms + 3 O22 molecules molecules 2 molecules of Al2 molecules of Al22OO33

4 moles of Al + 3 moles of O4 moles of Al + 3 moles of O22 2 moles of Al2 moles of Al22OO33

5

Chemical EquationsChemical Equations• Because the same atoms

are present in a reaction

at the beginning and at

the end, the amount of

matter in a system does

not change.

• The Law of the Law of the

Conservation of MatterConservation of Matter

6

Because of the principle of the

conservation of matterconservation of matter, an

equation must be equation must be

balancedbalanced..

It must have the same number of

atoms of the

same kind on both sides.

Chemical EquationsChemical Equations

Lavoisier, 1788

7

Balancing Balancing EquationsEquations

8

Balancing EquationsBalancing Equations

9

Balancing EquationsBalancing Equations

___C3H8 (g) + ___ O2 (g) ----> ___CO2 (g) + ___ H2O (g)

C3H8 (g) + 5 O2 (g) ----> 3 CO2 (g) + 4 H2O (g)

10

Balancing EquationsBalancing Equations___B___B44HH10 (g)10 (g) + ___ O + ___ O2 (g)2 (g) ---> ___ B ---> ___ B22OO3 (g)3 (g) + ___ H + ___ H22O O (g)(g)

2 B2 B44HH10 (g)10 (g) + 11 O + 11 O2 (g)2 (g) ---> 4 B ---> 4 B22OO3 (g)3 (g) + 10 H + 10 H22O O (g)(g)

STOICHIOMETRYSTOICHIOMETRY

- the study of the the study of the quantitativequantitative aspects of aspects of chemical chemical reactions.reactions.

11

12

STEP 1STEP 1Write the balanced chemical equationWrite the balanced chemical equation

NHNH44NONO33 N N22O + 2 HO + 2 H22OO

PROBLEM: If 454 g of NH4NO3 decomposes, how much H2O and N2O are formed? What is the theoretical yield of products?

13

454 g of NH4NO3 --> N2O + 2 H2O

STEP 2STEP 2

Convert reactant mass to molesConvert reactant mass to moles

(454 g) --> moles(454 g) --> moles

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3

14

454 g of NH4NO3 --> N2O + 2 H2O

STEP 3STEP 3 Convert moles reactant --> moles product.Convert moles reactant --> moles product.

Relate moles NHRelate moles NH44NONO33 to moles product. to moles product.

1 mol NH1 mol NH44NONO33 --> 2 mol H --> 2 mol H22OOExpress this relation as theExpress this relation as the

STOICHIOMETRIC FACTORSTOICHIOMETRIC FACTOR

2 mol H2O produced1 mol NH4NO3 used

15

454 g of NH4NO3 --> N2O + 2 H2O

= 11.4 mol H= 11.4 mol H22O producedO produced

STEP 3STEP 3

Convert moles reactant (5.68 mol) Convert moles reactant (5.68 mol) moles productmoles product

5.68 mol NH5.68 mol NH 44NONO33 •• 22 mol Hmol H22OO producedproduced

1 mol NH1 mol NH44NONO33 usedused

16

454 g of NH4NO3 --> N2O + 2 H2O

STEP 4STEP 4 Convert moles product (11.4 mol) to mass

product.

This is called the

THEORETICAL YIELDTHEORETICAL YIELD

This is the “Expected” # This is the “Expected” # of moles.of moles.

17

454 g of NH4NO3 --> N2O + 2 H2O

This is the “Expected” Mass!This is the “Expected” Mass!

ALWAYS FOLLOW THESE STEPS IN SOLVING ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!STOICHIOMETRY PROBLEMS!

Repeat to find the grams of N2O formed.

STEP 4STEP 4 Convert moles prod. (11.4 mol) to mass prod.Convert moles prod. (11.4 mol) to mass prod.

18.02 gO 11.4 mol H2O

•

1 mol = 204 g H2

18

454 g of NH4NO3 --> N2O + 2 H2O

STEP 5 STEP 5 How much NHow much N22O is formed?O is formed?

Total mass of reactants=total mass of Total mass of reactants=total mass of

productsproducts

454 g NH454 g NH44NONO33 = ___ g N = ___ g N22O + 204 g HO + 204 g H22OO

mass of Nmass of N22O = O = 250. g250. g

This is an alternate method.This is an alternate method.

19

454 g of NH4NO3 --> N2O + 2 H2O

STEP 6 STEP 6

Calculate the percent yield.Calculate the percent yield.

If you isolated only 131 g of NIf you isolated only 131 g of N22O, what is the O, what is the

percent yield?percent yield?

This compares the theoretical (250. g) and This compares the theoretical (250. g) and

actual (131 g) yields.actual (131 g) yields.

20

454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O

STEP 6STEP 6

Calculate the percent yield.Calculate the percent yield.

%% yield yield == actual yieldactual yield

theoretical yieldtheoretical yield •• 100%100%

%% yield yield == 131 g131 g

250. g250. g •• 100% 100% == 52.4%52.4%

21

Mass reactant

Molesreactant

Moles product

Mass product

General Plan For General Plan For Stoichiometry CalculationsStoichiometry Calculations

22

PROBLEM: Using 5.00 g of HPROBLEM: Using 5.00 g of H22OO22, what mass of , what mass of OO22 and of H and of H22O can be obtained?O can be obtained?

2 H2 H22OO22(liq)(liq) ---> ---> 2 H 2 H22OO(g)(g) + O + O22(g)(g)

Reaction is catalyzed by MnOReaction is catalyzed by MnO22

Step 1: moles of HStep 1: moles of H22OO22

Step 2: use STOICHIOMETRIC FACTOR to Step 2: use STOICHIOMETRIC FACTOR to calculate moles of Ocalculate moles of O22

Step 3: mass of OStep 3: mass of O22

Repeat for HRepeat for H22O.O.

Reactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANT

• In a given reaction, there is not In a given reaction, there is not enough of one reagent to use up the enough of one reagent to use up the other reagent completely.other reagent completely.

• The reagent in short supply The reagent in short supply LIMITSLIMITS thethe quantity of product that can be quantity of product that can be formed.formed.

24

LIMITING REACTANTSLIMITING REACTANTS

Reactants Products

2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________ Excess reactant = ____________

NOO2

25

LIMITING LIMITING

REACTANTSREACTANTS

26

Rxn 1Rxn 1 Rxn 2Rxn 2 Rxn 3Rxn 3

mass Zn (g)mass Zn (g) 7.007.00 3.273.27 1.311.31

mol Znmol Zn 0.1070.107 0.0500.050 0.0200.020

mol HClmol HCl 0.1000.100 0.1000.100 0.1000.100

mol HCl/mol Zn mol HCl/mol Zn 0.930.93 2.002.00 5.005.00

LIMITING REACTANTSLIMITING REACTANTS

React solid Zn with 0.100 mol HCl (aq)

Zn + 2 HCl ---> ZnCl2 + H2

27

PROBLEM: Mix 5.40 g of Al with 8.10 g PROBLEM: Mix 5.40 g of Al with 8.10 g of Clof Cl22. How many grams of Al. How many grams of Al22ClCl66 can can form?form?

28

Step 1 of LR problem: Step 1 of LR problem: compare actual mole compare actual mole ratio of reactants to ratio of reactants to theoretical mole ratio.theoretical mole ratio.

29

2 Al + 3 Cl2 ---> Al2Cl6

Reactants must be in the mole ratioReactants must be in the mole ratio

Step 1 of LR problem: compare Step 1 of LR problem: compare actual mole ratio of reactants to actual mole ratio of reactants to theoretical mole ratio.theoretical mole ratio.

mol Cl2mol Al

= 32

30

Deciding on the Limiting ReactantDeciding on the Limiting Reactant

IfIf

then there is not enough Al then there is not enough Al to use up all the Clto use up all the Cl22, and the , and the

limiting reagent islimiting reagent is

AlAl..

2 Al + 3 Cl2 Al + 3 Cl22 AlAl22ClCl66

molmol ClCl22mol Almol Al

>> 33

22

31

Deciding on the Limiting ReactantDeciding on the Limiting Reactant

IfIf

then there is not enough Clthen there is not enough Cl22

to use up all the Al, and the to use up all the Al, and the

limiting reagent islimiting reagent is

2 Al + 3 Cl2 Al + 3 Cl22 --->---> Al Al22ClCl66

ClCl22


<< 33

22

32

We have 5.40 g of Al and 8.10 g of ClWe have 5.40 g of Al and 8.10 g of Cl22

Step 2Step 2 of LR problem: Calculate of LR problem: Calculate moles of each reactantmoles of each reactant

55..4040 g Al g Al •• 1 mol1 mol

27.0 g27.0 g == 0.200 mol Al0.200 mol Al

88.10 g Cl.10 g Cl 22 •• 1 mol1 mol

70.9 g70.9 g == 0.114 mol Cl0.114 mol Cl 22

33

Find mole ratio of reactantsFind mole ratio of reactants

This This should be 3/2 or 1.5/1 if should be 3/2 or 1.5/1 if reactants are present in the reactants are present in the exact stoichiometric ratio.exact stoichiometric ratio.

Limiting reagent is Limiting reagent is ClCl22


== 0.114 mol 0.114 mol

0.200 mol 0.200 mol == 0.570.57

34

Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22. What . What mass of Almass of Al22ClCl66 can form? can form?

Limiting reactant = ClLimiting reactant = Cl22

Base all calculations on ClBase all calculations on Cl22

molesmolesClCl22

moles moles AlAl22ClCl66

gramsgramsClCl22

grams grams AlAl22ClCl66

1 mol Al2Cl63 mol Cl2

35

CALCULATIONS: calculate mass ofCALCULATIONS: calculate mass ofAlAl22ClCl66 expected. expected.

Step 1: Calculate moles of AlStep 1: Calculate moles of Al22ClCl66

expected based on LR.expected based on LR.

00..114114 molmol ClCl22 •• 11 molmol AlAl 22ClCl 66

3 mol Cl3 mol Cl22 == 0.0380 mol Al0.0380 mol Al22ClCl 66

Step 2: Calculate mass of AlStep 2: Calculate mass of Al22ClCl66

expected based on LR.expected based on LR.

00..03800380 molmol AlAl22ClCl66 •• 2266.4 g Al66.4 g Al22ClCl66

molmol == 10.1 g Al10.1 g Al22ClCl66

36

• ClCl22 was the limiting reactant. was the limiting reactant.

Therefore, Al was present in excess. Therefore, Al was present in excess.

But how much?But how much?

• First find how much Al was required.First find how much Al was required.

• Then find how much Al Then find how much Al is in excess.is in excess.

How much of which reactant will How much of which reactant will remain when reaction is complete?remain when reaction is complete?

37

2 Al + 3 Cl2 productsproducts

0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR

Calculating Excess AlCalculating Excess Al

38

2 Al + 3 Cl2 Al + 3 Cl22 productsproducts

0.200 mol0.200 mol 0.114 mol = LR0.114 mol = LR

Calculating Excess AlCalculating Excess Al

Excess Al = Al available - Al requiredExcess Al = Al available - Al required

= 0.200 mol - 0.0760 mol = 0.200 mol - 0.0760 mol

= 0.124 mol Al in excess= 0.124 mol Al in excess

00.114 mol Cl.114 mol Cl22 •• 2 mol Al2 mol Al

3 mol Cl3 mol Cl22 == 0.0760 mol Al req'0.0760 mol Al req'dd

39

N2 + 3 I2 2 NI3

Nitrogen and iodine react to form Nitrogen and iodine react to form nitrogennitrogen tritri iodide. If 50.0 g of nitrogen is iodide. If 50.0 g of nitrogen is mixed with 350.0 g iodine, calculate the mixed with 350.0 g iodine, calculate the number of grams of product formed and number of grams of product formed and the grams of reactant remaining.the grams of reactant remaining.

28.0 g/mol 253.8 g/mol 394.7 g/mol

1.79 mole 1.38 mole

L.R. (0.460 S.U.)

50.0g -12.9g 0g left 363 g

37.1g left 350.0g + 50.0g = 400.0g = 363g + 37.1g

Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula

Burn 0.115 g of a hydrocarbon, CBurn 0.115 g of a hydrocarbon, CxxHHyy, and produce , and produce

0.379 g of CO0.379 g of CO22 and 0.1035 g of H and 0.1035 g of H22O. O.

What is the empirical formula of CWhat is the empirical formula of CxxHHyy??

CCxxHHy y + oxygen 0.379 g CO + oxygen 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

4040

41

Chemical Molecular AnalysisChemical Molecular Analysis in the labin the lab

42


First, recognize that all C in CO2 and all H in H2O

is from CxHy.

1. Calculate moles of C in CO2

8.61 x 10-3 mol C

2. Calculate moles of H in H2O

1.149 x 10 -2 mol H

CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O

43


Now find ratio of mol H/mol C to find values Now find ratio of mol H/mol C to find values of x and y in Cof x and y in CxxHHyy..

1.149 x 10 1.149 x 10 -2 -2 mol H/ 8.61 x 10mol H/ 8.61 x 10-3 -3 mol C mol C

= 1.33 mol H / 1.00 mol C= 1.33 mol H / 1.00 mol C

= 4 mol H / 3 mol C= 4 mol H / 3 mol C

Empirical formula = CEmpirical formula = C33HH44

CCxxHHy y + some oxygen + some oxygen 0.379 g CO 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

44

Chemical AnalysisChemical Analysis CombustionCombustion--Determine a FormulaDetermine a Formula

Recognize that all C in CORecognize that all C in CO22 and all H in H and all H in H22O is from CO is from CxxHHyy..

1.1. Calculate moles of C in COCalculate moles of C in CO22

CCxxHHy y + some oxygen + some oxygen 0.379 g CO 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

0.379 g CO0.379 g CO22 mole CO mole CO22 mole C mole C

44.0 g CO44.0 g CO22 mole CO mole CO22

= 0.00861 mole C

45


2. Calculate moles of H in H2. Calculate moles of H in H22OO

CCxxHHyy+some oxygen 0.379gCO+some oxygen 0.379gCO22+0.1035 g H+0.1035 g H22OO

0.1035 g H2O mole H2O 2 mole H

18.0 g H2O mole H2O= 0.0115 mole H

46


Now find ratio of mol H/mol C to find values Now find ratio of mol H/mol C to find values of x and y in Cof x and y in CxxHHyy..

0.01150.0115 mol H/ 0.00861 mol H/ 0.00861 mol C mol C



Empirical formula = CEmpirical formula = C33HH44

CCxxHHy y + some oxygen 0.379 g CO + some oxygen 0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO

47Sample ProblemsSample Problems

1) A 0.537 g sample of an unknown compound 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of COburned to produced 1.030 g of CO22 and 0.632 g of H and 0.632 g of H22O. O.

Determine the empirical formula. Given that the Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, molecular weight is approximately 90 g/mole, determine the molecular formula.determine the molecular formula.

1.030 g CO2 mole CO2 mole C 12.0 g C 44.0g CO2 mole CO2 mole C

= 0.281 g C

0.632 g H2O mole H2O 2 mole H 1.0 g H 18.0 g H2O mole H2O mole H

= 0.070 g H


1) A 0.537 g sample of an unknown compound 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of COburned to produced 1.030 g of CO22 and 0.632 g of H and 0.632 g of H22O. O.

Determine the empirical formula. Given that the Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, molecular weight is approximately 90 g/mole, determine the molecular formula.determine the molecular formula.

- 0.281 g C- 0.281 g C

- 0.070 g H- 0.070 g H

0.537 g C, H, O0.537 g C, H, O

0.186 g O0.186 g O

49

..281 g 1 mole

12.0 g

.070 g 1 mole

1.0 g

C H O

2.02 6.0 1.00

Empirical formulaEmpirical formula C C22HH66OO

..186 g 1 mole

16.0 g

0.0234 0.070 0.0116

0.0116 0.0116 0.0116

Sample ProblemsSample Problems


Empirical formula Empirical formula CC22HH66O O

90

46= 2

Molecular formula Molecular formula CC44HH1212OO22

Alternate methodAlternate method. Use grams of cmpd. and . Use grams of cmpd. and its molar mass to determine moles of cmpd. its molar mass to determine moles of cmpd. Divide moles of C and H by these moles to find Divide moles of C and H by these moles to find the subscripts for C and H. The subscript for O the subscripts for C and H. The subscript for O can be determinded by difference.can be determinded by difference.


2) A 0.1247 g sample of ascorbic acid, vitamin C, was 2) A 0.1247 g sample of ascorbic acid, vitamin C, was burned to produce 0.1869 g of COburned to produce 0.1869 g of CO22 and 0.0510 g H and 0.0510 g H22O. O.

Ascorbic acid contains only carbon, hydrogen, and Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 Given that the molecular weight is approximately 180 g/mole, determine the molecular formula.g/mole, determine the molecular formula.

.1869 g CO2 mole CO2 mole C 12.0 g C 44.0 g CO2 mole CO2 mole C

= 0.0510 g C0.0510 g C

.0510 g H2O mole H2O 2 mole H 1.0 g H 18.0 g H2O mole H2O mole H

= 0.0057 g H 0.0057 g H


2) A 0.1247 g sample of ascorbic acid, vitamin C, was 2) A 0.1247 g sample of ascorbic acid, vitamin C, was burned to produce 0.1869 g of COburned to produce 0.1869 g of CO22 and 0.0510 g H and 0.0510 g H22O. O.

Ascorbic acid contains only carbon, hydrogen, and Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, the molecular weight is approximately 180 g/mole, determine the molecular formula.determine the molecular formula.

- 0.0510 g C

- 0.0057 g H

0.1247 g C, H, O

0.0680 g O0.0680 g O

53

..0510g 1 mole

12.0 g

.0057g 1 mole

1.0 g

C H O

1.00 1.3 1.00

Empirical formulaEmpirical formula CC33HH44OO33

..0680g 1 mole

16.0 g

.00425 .0057 .00425

.00425 .00425 .00425

Sample ProblemsSample Problems

3.00 3.9 3.00


Empirical formula Empirical formula CC33HH44OO33

180

88= 2

Molecular formula Molecular formula CC66HH88OO66

55

Chemical Analysis Chemical Analysis MixturesMixtures--Determine a PercentDetermine a Percent

1. The amount of calcium present in milk can be 1. The amount of calcium present in milk can be determined by adding oxalate ion, Cdetermined by adding oxalate ion, C22OO44

2-2-(in the (in the

form of its water-soluble sodium salt, Naform of its water-soluble sodium salt, Na22CC22OO44); the ); the

insoluble compound calcium oxalate is insoluble compound calcium oxalate is precipitated. Suppose you take a 75.0 g sample of precipitated. Suppose you take a 75.0 g sample of milk and isolate 0.288 g of calcium oxalate from it. milk and isolate 0.288 g of calcium oxalate from it. What is the weight percentage of calcium in the What is the weight percentage of calcium in the milk?milk?

MILK (Ca2+) CaC2O4

Na2C2O4

56


MILK (Ca2+) CaC2O4

75.0 g 0.288 g

? % Ca

Na2C2O4

% Ca = g Ca

75.0 g milk X 100

57


MILK (Ca2+) CaC2O4

75.0 g 0.288 g

? % Ca

Na2C2O4

% Ca = 0.0902 g Ca

75.0 g milk X 100

.288 g CaC2O4 mole CaC2O4 mole Ca 40.1 g Ca

128.1 g CaC2O4 mole CaC2O4 mole Ca = 0.0902 g Ca

= 0.120%

58


2. A 4.22 g sample of calcium chloride and 2. A 4.22 g sample of calcium chloride and sodium chloride was dissolved in water, sodium chloride was dissolved in water, and the solution was treated with sodium and the solution was treated with sodium carbonate to precipitate the calcium as carbonate to precipitate the calcium as calcium carbonate. After isolating the solid calcium carbonate. After isolating the solid calcium carbonate, it was heated to drive calcium carbonate, it was heated to drive off the carbon dioxide and form 0.959 g of off the carbon dioxide and form 0.959 g of calcium oxide. What is the weight percent calcium oxide. What is the weight percent of calcium chloride in the original 4.22 g of calcium chloride in the original 4.22 g sample?sample?

59


CaCl2/NaCl CaCO3

Na2CO3

CaO

Heat

0.959 g0.959 g

4.22 g

% CaCl2 = g CaCl2

4.22 g sample X 100

60


0.959 g CaO mole CaO mole CaCl2 111.1 g CaCl2

56.1 g CaO mole CaO 4 mole CaCl2

= 1.90 g CaCl2

= 45.0% % CaCl2 = 1.90 g CaCl2

4.22 g sample X 100

61

Practice ProblemsPractice Problems

1. Balance the following equations:

CS2 + Cl2 --> CCl4 + S2Cl2

N2 + O2 --> NO

C8H18 + O2 --> CO2 + H2O

2. Write the formula equation for each of the following:

sodium + water --> sodium hydroxide + hydrogen

magnesium + oxygen --> magnesium oxide

aluminum + hydrochloric acid -->

aluminum chloride + hydrogen

62


2. (continue)2. (continue)

aluminum + hydrochloric acid --> aluminum + hydrochloric acid -->

aluminum chloride + hydrogenaluminum chloride + hydrogen

sodium chlorate --> sodium chloride + oxygensodium chlorate --> sodium chloride + oxygen

mercury(II) sulfate + ammonium sulfide --> mercury(II) sulfate + ammonium sulfide -->

mercury(II) sulfide + ammonium sulfatemercury(II) sulfide + ammonium sulfate

iron + cupric sulfate --> iron(III) sulfate + copperiron + cupric sulfate --> iron(III) sulfate + copper

63


For problems 3-6For problems 3-6

3 H3 H22 + N + N22 2 NH 2 NH33

3. How many moles of H3. How many moles of H22 are required to are required to react 4.2 moles of Nreact 4.2 moles of N22??

4. How many moles of H4. How many moles of H22 are required to react are required to react 74 grams of N 74 grams of N22??

5. How many grams of NH5. How many grams of NH33 would be would be produced from 45 g of Hproduced from 45 g of H22??

6. How many moles of NH6. How many moles of NH3 3 would be would be produced from the reaction of 18.5 g Hproduced from the reaction of 18.5 g H22 and and 95 g of N95 g of N22??

64

Practice ProblemsPractice Problems7. Phosphine, PH3, is formed when calcium phosphide is added to

water. How many grams of phosphine can be obtained from 205 g of calcium phosphide?

8. How many grams of iron will be required to release all of the antimony from 10.0 g antimony trisulfide? (Ferrous sulfide is formed)

9. If calcium oxide were prepared by heating calcium carbonate, how many grams of the carbonate would be required to produce the 15.0 g of the oxide?

65

Practice ProblemsPractice Problems10. How many grams of cupric sulfate are needed to

completely react with 145 g of sodium chloride? How many grams of sodium sulfate could be produced by this reaction?

11. How many grams of sulfuric acid will react with 40.0 g of aluminum metal?

12. 75 grams of zinc are added to 120 grams of nitrous acid. How many grams of hydrogen gas are evolved?

66

Practice ProblemsPractice Problems13. 10.0 grams of hydrogen and 75 grams of oxygen are

exploded together in a reaction tube. How many grams of water are produced? What other gas is found in the tube(besides water vapor) after the reaction, and how many grams of this gas are there?

67

Practice ProblemsPractice Problems14. A white powder was a mixture of NaBr and

NaNO3. A sample of the powder weighing 1.341 g was dissolved in water and a solution of AgNO3 was added until the precipitation of AgBr was complete. The reaction mixture was filtered and dried and the precipitate of AgBr weighed 1.896 g. What was the percentage by weight of NaBr in the original sample?

68

Practice ProblemsPractice Problems15. A 4.81 g sample of an unknown compound

containing only carbon, hydrogen, and nitrogen was burned to produce 13.74 g of CO2 and 1.68 g of H2O. Determine the empirical formula.

69

Practice Problems AnswersPractice Problems Answers

1. 1,3,1,1 1,1,2 2,25,16,18

2. 2 Na + 2 HOH --> 2 NaOH + H2

2 Mg + O2 --> 2 MgO

2 Al + 6 HCl --> 2 AlCl3 + 3 H2

2 NaClO3 --> 2 NaCl + 3 O2

HgSO4 + (NH4)2S --> Hg + (NH4)2SO4

2 Fe + 3 CuSO4 --> Fe2(SO4)3 + 3 Cu

3. 13 mole H2 4. 7.9 mole H2

5. 260 g NH3 6. 6.2 mole NH3

70

Practice Problems AnswersPractice Problems Answers

7. 76.5 g PH3

8. 7.68 g Fe

9. 26.8 g CaCO3

10. 198 g CuSO4, 176 g Na2SO4

11. 218 g H2SO4

12. 2.3 g H2

13. 84 g H2O, 0.6 g H2

14. 77.48%

15. C5H3N

71

Writing EquationsWriting Equations

zinc + chlorine ---> zinc chloridezinc + chlorine ---> zinc chloride

Zn (s) + ClZn (s) + Cl22 (g) --> ZnCl (g) --> ZnCl22 (s) (s)

Combination, SynthesisCombination, Synthesis

72


potassium nitrate --> potassium nitrite + oxygenpotassium nitrate --> potassium nitrite + oxygen

KNOKNO3 3 (s) --> KNO(s) --> KNO22 (s) + O (s) + O2 2 (g)(g)

2 KNO2 KNO3 3 (s) --> 2 KNO(s) --> 2 KNO22 (s) + O (s) + O2 2 (g)(g)

DecompositionDecomposition

73


magnesium bromide + chlorine --> magnesium chloride + brominemagnesium bromide + chlorine --> magnesium chloride + bromine

MgBrMgBr22 (s) + Cl (s) + Cl22 (g) --> MgCl (g) --> MgCl22 (s) + Br (s) + Br22 (g) (g)

Single DisplacementSingle Displacement

74


calcium hydroxide + hydrochloric acid --> calcium hydroxide + hydrochloric acid --> calcium chloride + watercalcium chloride + water

Ca(OH)Ca(OH)22 (aq) + HCl (aq) --> CaCl (aq) + HCl (aq) --> CaCl22 (aq) + H (aq) + H22O (l)O (l)

Ca(OH)Ca(OH)22 (aq) + 2HCl (aq) --> CaCl (aq) + 2HCl (aq) --> CaCl22 (aq) + 2 H (aq) + 2 H22O (l)O (l)

Double DisplacementDouble Displacement

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zinc chloride + ammonium sulfide --> zinc chloride + ammonium sulfide --> zinc sulfide + ammonium chloride zinc sulfide + ammonium chloride

ZnClZnCl22 (aq) + (NH (aq) + (NH44))22S (aq) --> ZnS (s) + NHS (aq) --> ZnS (s) + NH44Cl (aq)Cl (aq)

ZnClZnCl22 (aq) + (NH (aq) + (NH44))22S (aq) --> ZnS (s) + 2 NHS (aq) --> ZnS (s) + 2 NH44Cl (aq)Cl (aq)

76


aluminum + cupric chloride -->aluminum + cupric chloride --> copper + aluminum chloride copper + aluminum chloride

Al (s) + CuClAl (s) + CuCl22 (aq) --> Cu (s) + AlCl (aq) --> Cu (s) + AlCl33 (aq) (aq)

2 Al (s) + 3 CuCl2 Al (s) + 3 CuCl22 (aq) --> 3 Cu (s) + 2 AlCl (aq) --> 3 Cu (s) + 2 AlCl33 (aq) (aq)

77

PROBLEM: PROBLEM: How many moles of HHow many moles of H22 are required to produce 9 moles of are required to produce 9 moles of NHNH33??


3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33

78

STEP 2STEP 2

Write the given and requested information Write the given and requested information below the equation.below the equation.

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33

? mole? mole 9 mole 9 mole

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3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 ? ? molemole 9 mole 9 mole

STEP 3STEP 3

Calculate using the information.Calculate using the information.

= 10 mole H= 10 mole H22

9 mole NH9 mole NH33 3 mole H 3 mole H22

2 mole NH2 mole NH33

80

PROBLEM: How many moles of NH3 can be produced from 10.4 moles of N2?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 10.4 mole 10.4 mole ? mole? mole

= 20.8 mole NH= 20.8 mole NH3310.4 mole N10.4 mole N22 2 mole NH 2 mole NH33

mole Nmole N22

81

PROBLEM: How many grams of H2 are required to produce 8 moles of NH3?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 ? g ? g8 mole8 mole

8 mole NH8 mole NH33 3 mole H 3 mole H22 2.0 g H 2.0 g H22

2 mole NH2 mole NH33 mole H mole H22

= 20 g H= 20 g H22

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PROBLEM: How many moles of NH3 can be produced from 55 grams of N2?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 55 g 55 g ? mole? mole

55 g N55 g N22 1 mole N 1 mole N22 2 mole NH 2 mole NH33

28.0 g N28.0 g N22 mole N mole N22

= 3.9 mole = 3.9 mole NHNH33

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PROBLEM: How many grams of H2 are required to react 24 grams of N2?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 ? g ? g 24 g 24 g

24 g N24 g N22 mole N mole N22 3 mole H 3 mole H22 2.0 g H 2.0 g H22

28.0 g N28.0 g N22 mole N mole N22 mole H mole H22

= 5.1 g H= 5.1 g H22

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PROBLEM: How many grams of N2 are required to produce 155 grams of NH3?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 ? g ? g 155 g 155 g

155 g NH155 g NH33 mole NH mole NH33 mole N mole N22 28.0 g N 28.0 g N22

17.0 g NH17.0 g NH33 2 mole NH 2 mole NH33 mole N mole N22

= 128 g N= 128 g N22

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Sample ProblemsSample Problems1) Sulfur dioxide may be oxidized to sulfur trioxide. 1) Sulfur dioxide may be oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted How many grams of sulfur dioxide could be converted by this process if 100.0 g of oxygen are available for by this process if 100.0 g of oxygen are available for the oxidation?the oxidation?

2 SO2 SO2 2 + O+ O22 --> 2 SO --> 2 SO33

? g? g 100.0 g 100.0 g

100.0 g O100.0 g O22 mole O mole O22 2 mole 2 mole SOSO22 64.1 g 64.1 g SOSO22

32.0 g O32.0 g O22 mole O mole O22 mole mole SOSO22

= 401 g = 401 g SOSO22

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Sample ProblemsSample Problems2) Lightning discharges in the atmosphere catalyze the 2) Lightning discharges in the atmosphere catalyze the conversion of nitrogen to nitrogen dioxide. How many conversion of nitrogen to nitrogen dioxide. How many grams of nitrogen would be required to make 25.0 g of grams of nitrogen would be required to make 25.0 g of nitrogen dioxide in this way?nitrogen dioxide in this way?

NN2 2 + 2 O+ 2 O22 --> 2 NO --> 2 NO22

? g? g 25.0 g 25.0 g

25.0 g NO25.0 g NO22 mole NO mole NO22 1 mole N 1 mole N22 28.0 g N 28.0 g N22

46.0 g NO46.0 g NO22 2 mole NO 2 mole NO22 mole N mole N22

= 7.61 g N= 7.61 g N22

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Sample ProblemsSample Problems3) Ferric oxide may be reduced to pure iron 3) Ferric oxide may be reduced to pure iron with coke (pure carbon). Suppose that 150.0 g with coke (pure carbon). Suppose that 150.0 g of ferric oxide is available. How many grams of of ferric oxide is available. How many grams of carbon would be needed? carbon would be needed?

2 Fe2 Fe22OO3 3 + 3 C + 3 C 4 Fe + 3 CO 4 Fe + 3 CO22 150.0 150.0 gg

? g? g

150.0 g Fe150.0 g Fe22OO33 mole Fe mole Fe22OO33 3 mole C 12.0 g C 3 mole C 12.0 g C

159.6 g Fe159.6 g Fe22OO33 2 mole Fe 2 mole Fe22OO3 3 mole C mole C

= 16.9 g C= 16.9 g C

88

Sample ProblemsSample Problems4) Zinc metal will react with hydrochloric acid to produce 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas react with all of the zinc? How many grams of hydrogen gas would be produced?would be produced?

ZnZn + 2 HCl --> ZnCl+ 2 HCl --> ZnCl22 + H + H22

50.0 g50.0 g ? g ? g ? g ? g

50.0 g Zn mole Zn 2 mole HCl 36.5 g HCl 50.0 g Zn mole Zn 2 mole HCl 36.5 g HCl

65.4 g Zn mole Zn mole HCl65.4 g Zn mole Zn mole HCl = 55.8gHCl= 55.8gHCl

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Sample ProblemsSample Problems4) Zinc metal will react with hydrochloric acid to produce 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas react with all of the zinc? How many grams of hydrogen gas would be produced?would be produced?

ZnZn + 2 HCl --> ZnCl+ 2 HCl --> ZnCl22 + H + H22

50.0 g ? g ? g 50.0 g ? g ? g

50.0 g Zn mole Zn mole H50.0 g Zn mole Zn mole H22 2.0 g H 2.0 g H22

65.4 g Zn mole Zn mole H65.4 g Zn mole Zn mole H22

= 1.5 g H= 1.5 g H22

90

Sample ProblemsSample Problems5) Phosphoric acid, H5) Phosphoric acid, H33POPO44, is produced in the reaction , is produced in the reaction

between calcium phosphate and sulfuric acid. How many between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in of calcium phosphate? What other product is formed and in what quantity?what quantity?

CaCa33(PO(PO44))2 2 + 3 H+ 3 H22SOSO4 4 --> 3 CaSO--> 3 CaSO44 + 2 H + 2 H33POPO44 55 g 55 g

? g ? g ? g ? g

55 g Ca55 g Ca33(PO4)(PO4)2 2 mole Ca mole Ca33(PO4)(PO4)22 3 mole CaSO 3 mole CaSO44 136.2 g CaSO 136.2 g CaSO44

310.3 gCa310.3 gCa33(PO4)(PO4)22 mole Ca mole Ca33(PO4)(PO4)22 mole CaSO mole CaSO44

= 72 g CaSO= 72 g CaSO44

91

Sample ProblemsSample Problems5) Phosphoric acid, H3PO4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be producedproduced from 55 grams of calcium phosphate? What other product is formed and in what quantity?

CaCa33(PO(PO44))2 2 + 3 H+ 3 H22SOSO4 4 --> 3 CaSO--> 3 CaSO44 + 2 H + 2 H33POPO44 55 g 55 g

? g ? g ? g ? g

55 g Ca55 g Ca33(PO4)(PO4)2 2 mole Ca mole Ca33(PO4)(PO4)22 2 mole H 2 mole H33POPO44 98.0 g H 98.0 g H33POPO44

310.3 gCa310.3 gCa33(PO4)(PO4)22 mole Ca mole Ca33(PO4)(PO4)22 mole H mole H33POPO44

= 35 g H= 35 g H33POPO44

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Percent YieldPercent Yield

% yield = actual yield

theoretical yield • 100%

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PROBLEM: If 19.3 g H2 produces 78.5 g NH3, what is the percent yield?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33 19.3 19.3 gg ? g ? g

19.3 g H19.3 g H22 mole H mole H22 2 mole NH 2 mole NH33 17.0 g NH 17.0 g NH33

2.0 g H2.0 g H22 3 mole H 3 mole H2 2 mole NH mole NH33

= 110 g = 110 g NHNH33

% yield =% yield =78.5g78.5g

110 g110 gx 100 = 71 %x 100 = 71 %

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PROBLEM: If the yield obtained is 75%, how many grams of NH3 would be obtained from 10.4 g of N2?

3 H3 H2 2 + N+ N22 --> 2 NH --> 2 NH33

10.4 g10.4 g ? g ? g

10.4 g N10.4 g N22 mole N mole N22 2 mole NH 2 mole NH33 17.0 g NH 17.0 g NH33

28.0 g N28.0 g N22 mole N mole N2 2 mole NH mole NH33

= 12.6 g = 12.6 g

NHNH3312.6 g NH12.6 g NH33 x .75 = 9.4 g NH x .75 = 9.4 g NH33

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PROBLEM: 75.0 grams of potassium hydroxide are permitted to react with 50.0 grams of hydrochloric acid. How many grams of potassium chloride are formed?


KOHKOH + HCl --> KCl + HOH+ HCl --> KCl + HOH

96

STEP 2STEP 2

Write the given and requested information Write the given and requested information below the equation.below the equation.


75.0 g 50.0 g ?g75.0 g 50.0 g ?g

97


75.0 g 50.0 g ?g75.0 g 50.0 g ?g

STEP 3STEP 3 Calculate the product assuming

that each reactant is the limiting reagent.

75.0 g KOH mole KOH mole KCl 74.6 g KCl75.0 g KOH mole KOH mole KCl 74.6 g KCl

56.1 g KOH mole KOH mole KCl56.1 g KOH mole KOH mole KCl= 99.7g KCl= 99.7g KCl

50.0 g HCl mole HCl mole KCl 74.6 g KCl50.0 g HCl mole HCl mole KCl 74.6 g KCl

36.5 g HCl mole HCl mole KCl36.5 g HCl mole HCl mole KCl

= 102 g KCl= 102 g KCl

98


75.0 g 50.0 g ?g75.0 g 50.0 g ?g

STEP 4STEP 4 Determine the limiting

reactant and the actual amount of product.

75.0 g KOH can produce 99.7 g KCl75.0 g KOH can produce 99.7 g KCl

KOH is the limiting reactant, HCl is KOH is the limiting reactant, HCl is the excess reactant.the excess reactant.

50.0 g HCl can produce 102 g KCl50.0 g HCl can produce 102 g KCl

99.7 g KCl99.7 g KCl

99


75.0 g 50.0 g ?g75.0 g 50.0 g ?g

STEP 5STEP 5Determine the amount of excess

reactant by calculating the amount used and subtracting from the starting amount.

75.0 g KOH mole KOH mole HCl 36.5 g HCl75.0 g KOH mole KOH mole HCl 36.5 g HCl

56.1 g KOH mole KOH mole HCl56.1 g KOH mole KOH mole HCl= 48.8g HCl= 48.8g HCl usedused

50.0 g - 48.8 = 1.2 g HCl left50.0 g - 48.8 = 1.2 g HCl left


1) How many grams of carbon dioxide can be 1) How many grams of carbon dioxide can be obtained from the action of 100.0 grams of obtained from the action of 100.0 grams of hydrobromic acid on 100.0 grams of calcium hydrobromic acid on 100.0 grams of calcium carbonate?carbonate?

2 HBr + CaCO2 HBr + CaCO3 3 --> HOH + CO --> HOH + CO22 + CaBr + CaBr22

100.0 g 100.0 g ? g100.0 g 100.0 g ? g

100.0 g HBr mole HBr mole CO100.0 g HBr mole HBr mole CO22 44.0 g CO 44.0 g CO22

80.9 g HBr 2 mole HBr mole CO80.9 g HBr 2 mole HBr mole CO22

= 27.2 g CO= 27.2 g CO22

100.0 g CaCO100.0 g CaCO33 mole CaCO mole CaCO33 mole CO mole CO22 44.0 g CO 44.0 g CO22

100.1 g CaCO100.1 g CaCO33 mole CaCO mole CaCO33 mole CO mole CO22

= 44.0 g CO= 44.0 g CO22


2) How many grams of ammonia are evolved 2) How many grams of ammonia are evolved when 34 grams of ammonium chloride are when 34 grams of ammonium chloride are added to 140 grams of barium hydroxide?added to 140 grams of barium hydroxide?

2 NH2 NH44Cl + Ba(OH)Cl + Ba(OH)2 2 --> BaCl --> BaCl22 + 2 NH + 2 NH33 +2 HOH +2 HOH

34 g 140 g ? g34 g 140 g ? g

34 g NH34 g NH44Cl mole NHCl mole NH44Cl 2 mole NHCl 2 mole NH33 17.0 g NH 17.0 g NH33

53.5 g NH53.5 g NH44Cl 2 mole NHCl 2 mole NH44Cl mole NHCl mole NH33

= 11 g NH= 11 g NH33

140 g Ba(OH)140 g Ba(OH)22 mole Ba(OH) mole Ba(OH)22 2 mole NH 2 mole NH33 17.0 g NH 17.0 g NH33

171.3 g Ba(OH)171.3 g Ba(OH)22 mole Ba(OH) mole Ba(OH)22 mole NH mole NH33

= 28.0 gNH= 28.0 gNH33


3) 100.0 grams of lithium metal is dropped into 3) 100.0 grams of lithium metal is dropped into 1.000 liter of water. How many grams of 1.000 liter of water. How many grams of hydrogen are produced?hydrogen are produced?

2 Li + 2 HOH2 Li + 2 HOH --> 2 LiOH + H --> 2 LiOH + H22

100.0 g 1000. g ? g100.0 g 1000. g ? g

100.0 g Li mole Li mole H100.0 g Li mole Li mole H22 2.0 g H 2.0 g H2 2

6.9 g Li 2 mole Li mole H 6.9 g Li 2 mole Li mole H 22

= 14 g H= 14 g H22

1000. g HOH mole HOH mole H1000. g HOH mole HOH mole H22 2.0 g H 2.0 g H22 18.0 g HOH 2 mole HOH 18.0 g HOH 2 mole HOH mole Hmole H22

= 56 g H= 56 g H22

103Sample ProblemsSample Problems4) .50.0 grams of oxygen are available for the 4) .50.0 grams of oxygen are available for the combustion of 25.0 grams of carbon. How many grams combustion of 25.0 grams of carbon. How many grams in excess is the oxygen or carbon ?in excess is the oxygen or carbon ?

OO22 + C + C --> CO --> CO22

? g excess 50.0 g 25.0 g? g excess 50.0 g 25.0 g

50.0 g O50.0 g O22 mole O mole O22 mole C 12.0 g C mole C 12.0 g C

32.0 g O32.0 g O22 mole O mole O22 mole C mole C

= 18.8g C used= 18.8g C used

25.0 g - 18.8 = 6.2 g C left25.0 g - 18.8 = 6.2 g C left

104Sample ProblemsSample Problems5) 140.0 grams of sulfuric acid is added to 230.0 grams 5) 140.0 grams of sulfuric acid is added to 230.0 grams of barium peroxide. Which reactant is in excess and by of barium peroxide. Which reactant is in excess and by

how many grams?how many grams?

HH22SOSO4 4 + BaO+ BaO2 2 --> BaSO --> BaSO44 + H + H22OO22

g excess 140. g 230. g g excess 140. g 230. g

140.gH140.gH22SOSO44 mole H mole H22SOSO4 4 mole BaO mole BaO22 169.3 g BaO 169.3 g BaO22

98.1 g H 98.1 g H22SOSO44 mole H mole H22SOSO4 4

mole BaOmole BaO2 2 = 242 g BaO= 242 g BaO22

Since only 230. g available, Since only 230. g available, BaOBaO22 is the limiting reactant is the limiting reactant

105Sample ProblemsSample Problems5) 140.0 grams of sulfuric acid is added to 230.0 grams 5) 140.0 grams of sulfuric acid is added to 230.0 grams of barium peroxide. Which reactant is in excess and by of barium peroxide. Which reactant is in excess and by

how many grams?how many grams?

HH22SOSO4 4 + BaO+ BaO2 2 --> BaSO --> BaSO44 + H + H22OO22

g excess 140. g 230. g g excess 140. g 230. g

230. g BaO230. g BaO2 2 mole BaO mole BaO22 mole H mole H22SOSO44 98.1 g H 98.1 g H22SOSO44

169.3 g BaO 169.3 g BaO22 mole BaO mole BaO22 mole H mole H22SOSO4 4

= 133 g H= 133 g H22SOSO4 4 usedused

140. g - 133 = 7 g H140. g - 133 = 7 g H22SOSO44 left left

chemical reactions chapter 4 reactants: zn + i 2 product: zni 2

Documents

o g slide

moles of o

o step

g of nh

o express

o problem

o mass of n

mol h mol h2 o