stoichiometrystoichiometry · 2 0 2 formula units mgo ... particles (atoms, molecules, formula...

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Solutions Manual Chemistry: Matter and Change • Chapter 11 209

StoichiometryStoichiometry

SOLUTIONS MANUALCHAPTER 11

Section 11.1 Defining Stoichiometrypages 368–372

Practice Problemspages 371–372

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

a. N2(g) � 3H2(g) → 2NH3(g)

1 molecule N2 � 3 moleculesH2 → 2 molecules NH3

1 mole N2 � 3 moles H2 → 2 moles NH3

Mass:

N2: (2 mol)(14.007 g/mol) � 28.014 g

3H2: (6 mol)(1.008 g/mol) � 6.048 g

2NH3: (2 mol)(14.007 g/mol) �(6 mol)(1.008 g/mol) � 34.062 g

28.014 g N2 � 6.048 g H2 0 34.062 g NH3

34.062 g reactants � 34.062 g products

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

1 molecule HCl � 1 formula unit KOH → 1 formula unit KCl � 1 molecule H2O

1 mole HCl � 1 mole KOH → 1 mole KCl � 1 mole H2O

Mass:

HCl: (1 mol)(1.008 g/mol) � (1 mol)(35.453 g/mol) � 36.461 g

KOH: (1 mol)(39.098 g/mol) � (1 mol)(15.999 g/mol) � (1 mol)(1.008 g/mol) � 56.105 g

KCl: (1 mol)(39.098 g/mol) � (1 mol)(35.453 g/mol) � 74.551 g

H2O: (2 mol)(1.008 g/mol) � (1 mol)( 15.999 g/mol) � 18.015 g

36.461 g HCl � 56.105 g KOH → 74.551 g KCl � 18.015 g H2O


c. 2Mg(s) � O2(g) → 2MgO(s)

2 atoms Mg � 1 moleculeO2 0 2 formula units MgO

2 moles Mg � 1 mole O2 → 2 moles MgO

Mass:

2Mg: (2 mol)(24.305 g/mol) � 48.610 g

O2: (2 mol)(15.999 g/mol) � 31.998 g

2MgO: (2 mol)(24.305 g/mol) � (2 mol)(15.999 g/mol) � 80.608 g

48.610 g Mg � 31.998 g O2 → 80.608 g MgO


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)

2Na(s) � 2H2O(l) → 2NaOH(aq) � 1H2(g)

2 atoms Na � 2 molecules H2O → 2 formula units NaOH � 1 molecule H2

2 mol Na � 2 mol H2O → 2 mol NaOH � 1 mol H2

Mass:

2Na: (2 mol)(22.990 g/mol) � 45.980 g

2H2O: (4 mol)(1.008 g/mol) � (2 mol)(15.999 g/mol) � 36.030 g

2NaOH: (2 mol)(22.990 g/mol) �(2 mol)(15.999 g/mol) � (2 mol)(1.008 g/mol) � 79.994 g

H2: (2 mol)(1.008 g/mol) � 2.016 g

45.980 g Na � 36.030 g H2O → 79.994 g NaOH � 2.016 g H2


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210 Chemistry: Matter and Change • Chapter 11 Solutions Manual


b. ___Zn(s) � ___HNO3(aq) → ___Zn(NO3)2(aq) � ___N2O(g) � ___H2O(l)

4Zn(s) � 10HNO3(aq) → 4Zn(NO3)2(aq) � 1N2O(g) � 5H2O(l)

4 atoms Zn � 10 molecules HNO3 → 4 formula units Zn(NO3)2 � 1 molecule N2O � 5 molecules H2O

4 mol Zn � 10 mol HNO3 → 4 mol Zn(NO3)2 � 1 mol N2O � 5 mol H2O

Mass:

4Zn: (4 mol)(65.39 g/mol) � 261.56 g

10HNO3: (10 mol)(1.008 g/mol) � (10 mol)(14.007 g/mol) � (30 mol)(15.999 g/mol) � 630.12 g

4Zn(NO3)2: (4 mol)(65.39 g/mol) � (8 mol)(14.007 g/mol) � (24 mol)(15.999 g/mol) � 757.592 g

N2O: (2 mol)(14.007 g/mol) � (1 mol)(15.999 g/mol) � 44.013 g

5H2O: (10 mol)(1.008 g/mol) � (5 mol)(15.999 g/mol) � 90.075 g

261.56 g Zn � 630.12 g HNO3 → 757.592 g Zn(NO3)2 � 44.013 g N2O � 90.075 g H2O


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

a. 4Al(s) � 3O2(g) → 2Al2O3(s)

4 mol Al _ 3 mol O2

3 mol O2 __

2 mol Al2O3

2 mol Al2O3 __ 4 mol Al

3 mol O2 _ 4 mol Al

2 mol Al2O3 __

3 mol O2 4 mol Al __

2 mol Al2O3

b. 3Fe(s) � 4H2O(l) → Fe3O4(s) � 4H2(g)

3 mol Fe __ 4 mol H2O

3 mol Fe _ 4 mol H2

3 mol Fe __ 1 mol Fe3O4

4 mol H2O

__ 3 mol Fe

4 mol H2 _ 3 mol Fe

1 mol Fe3O4 __

3 mol Fe

1 mol Fe3O4 __

4 mol H2

1 mol Fe3O4 __ 4 mol H2O

4 mol H2O

__ 4 mol H2

4 mol H2 __

1 mol Fe3O4

4 mol H2O __

1 mol Fe3O4

4 mol H2 __ 4 mol H2O

c. 2HgO(s) → 2Hg(l) � O2(g)

2 mol HgO

__ 2 mol Hg

1 mol O2 _ 2 mol Hg

1 mol O2 __ 2 mol HgO

2 mol Hg

__ 2 mol HgO

2 mol Hg

_ 1 mol O2

2 mol HgO

__ 1 mol O2

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

a. ZnO(s) � HCl(aq) → ZnCl2(aq) � H2O(l)

ZnO(s) � 2HCl(aq) → ZnCl2(aq) � H2O(l)

1 mol ZnO __ 2 mol HCl

1 mol ZnO __ 1 mol ZnCl2

1 mol ZnO __ 1 mol H2O

2 mol HCl __ 1 mol ZnO

2 mol HCl __ 1 mol ZnCl2

2 mol HCl __ 1 mol H2O

1 mol ZnCl2 __ 1 mol ZnO

1 mol ZnCl2 __

2 mol HCl

1 mol ZnCl2 __ 1 mol H2O

1 mol H2O

__ 1 mol ZnO

1 mol H2O

__ 2 mol HCl

1 mol H2O

__ 1 mol ZnCl2

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

2C4H10(g) � 13O2(g) 0 8CO2(g) � 10H2O(l)

2 mol C4H10 __ 13 mol O2

2 mol C4H10 __ 8 mol CO2

2 mol C4H10 __ 10 mol H2O

13 mol O2 __

2 mol C4H10

8 mol CO2 __ 2 mol C4H10

10 mol H2O

__ 2 mol C4H10

10 mol H2O

__ 13 mol O2

10 mol H2O

__ 8 mol CO2

8 mol CO2 _ 13 mol O2

13 mol O2 __

10 mol H2O

8 mol CO2 __ 10 mol H2O

13 mol O2 _ 8 mol CO2

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Section 11.1 Assessmentpage 372

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

The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. The masses of reactants and products are equal.

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

n � 3, thus (n)(n � 1) � (3)(2) � 6 mole ratios

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

particles (atoms, molecules, formula units), moles, and mass

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

xA/yB and xA/zAB

yB/xA and yB/zAB

zAB/xA and zAB/yB

9. Apply Hydrogen peroxide (H2O2) decomposes to produce water and oxygen. Write a balanced chemical equation for this reaction, and deter-mine the possible mole ratios.

2H2O2 → 2H2O � O2

2 mol H2O2/2 mol H2O, 2 mol H2O2/1 mol O2, 2 mol H2O/2 mol H2O2, 2 mol H2O/1 mol O2, 1 mol O2/2 mol H2O2, 1 mol O2/2 mol H2O

10. Model Write the mole ratios for the reaction of hydrogen gas and oxygen gas, 2H2(g) � O2(g) → 2H2O. Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Show the water molecules produced.

6H2 � 3O2 6H2O

2H2/O2 and 2H2/2H2O

O2/2H2 and O2/2H2O

2H2O/2H2 and 2H2O/O2

Sketches should show six hydrogen atoms combining with three oxygen atoms to form six water molecules.

Section 11.2 Stoichiometric Calculationspages 373–378

Practice Problemspages 375–377

11. Methane and sulfur react to produce carbon disulfide (CS2), a liquid often used in the production of cellophane.

___CH4(g) � ___S8(s) → ___CS2(l) � ___H2S(g)

a. Balance the equation.

2CH4(g) � S8(s) → 2CS2(l) � 4H2S(g)

b. Calculate the moles of CS2 produced when 1.50 moles of S8 is used.

1.50 mol S8 � 2 mol CS2 _ 1 mol S8

� 3.00 mol CS2

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c. How many moles of H2S are produced?

1.50 mol S8 � 4 mol H2S

_ 1 mol S8

� 6.00 mol H2S

12. Challenge Sulfuric acid (H2SO4) is formed when sulfur dioxide (SO2) reacts with oxygen and water.

a. Write the balanced chemical equation for the reaction.

2SO2(g) � O2(g) � 2H2O(l) → 2H2SO4(aq)

b. How many moles of H2SO4 are produced from 12.5 moles of SO2?

12.5 mol SO2 � 2 mol H2SO4 __

2 mol SO2

� 12.5 mol H2SO4 produced

c. How many moles of O2 are needed?

12.5 mol SO2 � 1 mol O2 _ 2 mol SO2

� 6.25 mol O2 needed

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

2.50 mol ? g

Electricenergy NaCl

Cl2

Na

Step 1: Balance the chemical equation.

2NaCl(s) → 2Na(s) � Cl2(g)

Step 2: Make mole → mole conversion.

2.50 mol NaCl � 1 mol Cl2 __

2 mol NaCl � 1.25 mol Cl2

Step 3: Make mole → mass conversion.

1.25 mol Cl2 � 70.9 g Cl2 _ 1 mol Cl2

� 88.6 g Cl2

14. Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4) is extracted from titanium oxide (TiO2) using chlorine and coke (carbon).

TiO2(s) � C(s) � 2Cl2(g) → TiCl4(s) � CO2(g)

a. What mass of Cl2 gas is needed to react with 1.25 mol of TiO2?


1.25 mol TiO2 � 2 mol Cl2 __

1 mol TiO2 � 2.50 mol Cl2


2.50 mol Cl2 � 70.90 g Cl2 __ 1 mol Cl2

� 177 g Cl2

b. What mass of C is needed to react with 1.25 mol of TiO2?


1.25 mol TiO2 � 1 mol C __ 1 mol TiO2

� 1.25 mol C


1.25 mol C � 12.011 g C

__ 1 mol C

� 15.0 g C

c. What is the mass of all of the products formed by reaction with 1.25 mol of TiO2?

Calculate the mass to TiO2 used.

1.25 mol TiO2 � 79.865 g TiO2 __

1 mol TiO2 � 99.8 g TiO2

Calculate the total mass of the reactants.

99.8 g � 177 g � 15.0 g � 292 g

Because mass is conserved, the mass of the products must equal the mass of reactants.

mass of products � 292 g

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15. One of the reactions used to inflateautomobile air bags involves sodium azide (NaN3): 2NaN3(s) → 2Na(s) � 3N2(g). Determine the mass of N2 produced from the decomposition of NaN3 shown below.

100.0 g NaN3 → ? g N2(g)

N2 gas

2NaN3(s) → 2Na(s) � 3N2(g)

Step 1: Make mass → mole conversion.

100.0 g NaN3 � 1 mol NaN3 __

65.02 g NaN3 � 1.538 mol NaN3


1.538 mol NaN3 � 3 mol N2 __

2 mol NaN3 � 2.307 mol N2


2.307 mol N2 � 28.02 g N2 __ 1 mol N2

� 64.64 g N2

16. Challenge In the formation of acid rain, sulfur dioxide (SO2) reacts with oxygen and water in the air to form sulfuric acid (H2SO4). Write the balanced chemical equation for the reac-tion. If 2.50 g of SO2 reacts with excess oxygen and water, how much H2SO4, in grams, is produced?

Step 1: Balance the chemical equation.

2SO2(g) � O2(g) � 2H2O(l) 0 2H2SO4(aq)

Step 2: Make mass → mole conversion.

2.50 g SO2 � 1 mol SO2 __

64.07 g SO2 � 0.0390 mol SO2


0.0390 mol SO2 � 2 mol H2SO4 __

2 mol SO2

� 0.0390 mol H2SO4


0.0390 mol H2SO4 � 98.09 g H2SO4 __ 1 mol H2SO4

� 3.83 g H2SO4


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

The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products.

18. List the four steps used in solving stoichio-metric problems.

1. Balance the equation.

2. Convert the mass of the known substance to moles of known substance.

3. Use the mole ratio to convert from moles of the known to moles of the unknown.

4. Convert moles of unknown to mass of the unknown.

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

moles of unknown/moles of known

20. Apply How can you determine the mass of liquid bromine (Br2) needed to react completely with a given mass of magnesium?

Write a balanced equation. Convert the given mass of magnesium to moles. Use the mole ratio from the balanced equation to convert moles of magnesium to moles of bromine. Convert from moles of bromine to mass of bromine.

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21. Calculate Hydrogen reacts with excess nitrogen as follows:

N2(g) � 3H2(g) → 2NH3(g)

If 2.70 g of H2 reacts, how many grams of NH3 is formed?

(2.70 g H2) ( 1 mol __ 2.016 g H2

) � 1.34 mol H2

1.34 mol H2 ( 2 mol NH3 _ 3 mol H2

) � 0.893 mol NH3

(0.893 mol NH3) ( 17.031 g NH3 __ 1 mol NH3

) � 15.2 g NH3

22. Design a concept map for the following reaction.

CaCO3(s) � 2HCl(aq) → CaCl2(aq) � H2O(l) � CO2(g)

The concept map should explain how to deter-mine the mass of CaCl2 produced from a given mass of HCl.

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.

Section 11.3 Limiting Reactantspages 379–384

Practice Problemspage 383

23. The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag: 6Na(s) � Fe2O3(s) → 3Na2O(s) � 2Fe(s). If 100.0 g of Na and 100.0 g of Fe2O3 are used in this reaction, determine the following.

a. limiting reactant

Make mass → mole conversion.

100.0 g Na � 1 mol Na __ 22.99 g Na

� 4.350 mol Na

100.0 g Fe2O3 � 1 mol Fe2O3 __

159.7 g Fe2O3

� 0.6261 mol Fe2O3

Make mole ratio comparison.

0.6261 mol Fe2O3 __

4.350 mol Na compared to

1 mol Fe2O3 __ 6 mol Na

0.1439 compared to 0.1667

The actual ratio is less than the needed ratio, so iron(III) oxide is the limiting reactant.

b. reactant in excess

Sodium is the excess reactant.

c. mass of solid iron produced

Make mole → mole conversion.

0.6261 mol Fe2O3 � 2 mol Fe __ 1 mol Fe2O3

� 1.252 mol Fe

Make mole → mass conversion.

1.252 mol Fe � 55.85 g Fe

_ 1 mol Fe

� 69.92 g Fe

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


0.6261 mol Fe2O3 � 6 mol Na __ 1 mol Fe2O3

� 3.757 mol Na needed


3.757 mol Na � 22.9 g Na

_ 1 mol Na

� 86.37 g Na needed

100.0 g Na given � 86.37 g Na needed� 13.6 g Na in excess

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.


6CO2(g) � 6H2O(l) 0 C6H12O6(aq) � 6O2(g)

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b. Determine the limiting reactant.


88.0 g CO2 � 1 mol CO2 __

44.01 g CO2 � 2.00 mol CO2

64.0 g H2O � 1 mol H2O

__ 18.02 g H2O

� 3.55 mol H2O

Make mole ratio comparison.

2.00 mol CO2 __ 3.55 mol H2O

compared to 6 mol CO2 __ 6 mol H2O

;

0.563 compared to 1.00

The actual ratio is less than the needed ratio, so carbon dioxide is the limiting reactant.

c. Determine the excess reactant.

Water is the excess reactant.

d. Determine the mass in excess.


2.00 mol CO2 � 6 mol H2O

__ 6 mol CO2

� 2.00 mol H2O


2.00 mol H2O � 18.02 g H2O

__ 1 mol H2O

� 36.0 g H2O needed

64.0 g H2O given � 36.0 g H2O needed � 28.0 g H2O in excess

e. Determine the mass of glucose produced.


2.00 mol CO2 � 1 mol C6H12O6 __

6 mol CO2

� 0.333 mol C6H12O6

Make mole 0 mass conversion.

0.333 mol C6H12O6 � 180.24 g C6H12O6 __

1 mol C6H12O6

� 60.0 g C6H12O6


25. Describe the reason why a reaction between two elements comes to an end.

one of the reactants is used up

26. Identify the limiting and the excess reactant in each reaction.

a. Wood burns in a campfire.

The wood limits. Oxygen is in excess. The fire will burn only while wood is present.

b. Airborne sulfur reacts with the silver plating on a teapot to produce tarnish (silver sulfide).

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. Baking powder in batter decomposes to produce carbon dioxide.

A decomposition reaction usually has only one reactant. The reaction is limited by the amount of baking powder present.

27. Analyze Tetraphosphorous trisulphide (P4S3) is used in the match heads of some matches. It is produced in the reaction 8P4 � 3S8 → 8P4S3. Determine which of the following statements are incorrect, and rewrite the incorrect state-ments to make them correct.

a. 4 mol P4 reacts with 1.5 mol S8 to form 4 mol P4S3.

correct

b. Sulfur is the limiting reactant when 4 mol P4 and 4 mol S8 react.

Phosphorus is the limiting reactant.

c. 6 mol P4 reacts with 6 mol S8, forming 1320 g P4S3.

correct

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Section 11.4 Percent Yieldpages 385–388

Practice Problemspage 387

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) → AlCl3(aq) � 3H2O(l). If 14.0 g of Al(OH)3 is present in an antacid tablet, deter-mine the theoretical yield of AlCl3 produced when the tablet reacts with HCl.


14.0 g Al(OH)3 � 1 mol Al(OH)3 __ 78.0 g Al(OH)3

� 0.179 mol Al(OH)3


0.179 mol Al(OH)3 � 1 mol AlCl3 __

1 mol Al(OH)3

� 0.179 mol AlCl3


0.179 mol AlCl3 � 133.3 g AlCl3 __ 1 mol AlCl3

� 23.9 g AlCl3

23.9 g of AlCl3 is the theoretical yield.

29. Zinc reacts with iodine in a synthesis reaction:

Zn � I2 0 ZnI2.

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

Write the balanced chemical equation.

Zn(s) � I2(s) → ZnI2(s)


1.912 mol Zn � 1 mol ZnI2 _ 1 mol Zn

� 1.912 mol ZnI2


1.912 mol ZnI2 � 319.2 g ZnI2 __ 1 mol ZnI2

� 610.3 g ZnI2

610.3 g of ZnI2 is the theoretical yield.

b. Determine the percent yield if 515.6 g of product is recovered.

% yield � 515.6 g ZnI2 __ 610.3 g ZnI2

� 100

� 84.48% yield of ZnI2

30. Challenge When copper wire is placed into a silver nitrate solution (AgNO3), silver crystals and copper(II) nitrate (Cu(NO3)2) solution form.


Cu(s) � 2AgNO3(aq) → 2Ag(s) � Cu(NO3)2(aq)

b. If a 20.0�g sample of copper is used, determine the theoretical yield of silver.


20.0 g Cu � 1 mol Cu __ 63.55 g Cu

� 0.315 mol Cu


0.315 mol Cu � 2 mol Ag

_ 1 mol Cu

� 0.630 mol Ag


0.630 mol Ag � 107.9 g Ag

__ 1 mol Ag

� 68.0 g Ag

68.0 g of Ag is the theoretical yield.

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

% yield � 60.0 g Ag

_ 68.0 g Ag

� 100

� 88.2% yield of Ag


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

percent yield

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

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. Explain how percent yield is calculated.

divide the actual yield by the theoretical yield and multiply the quotient by 100

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) → Fe2S3(s)

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

(83.77 g Fe) ( 1 mol Fe __ 55.845 g Fe

) � 1.500 mol Fe

1.500 mol Fe ( 1 mol Fe2S3 __ 2 mol Fe

) ( 207.885 g Fe2S3 __ 1 mol Fe2S3

) � 155.9 g Fe2S3

35. Calculate the percent yield of the reaction of magnesium with excess oxygen:

2Mg(s) � O2(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

mass of Mg � 38.06 g � 35.67 g � 2.39 g

mass of MgO � 39.15 g � 33.67 g � 3.48 g

Theoretical yield: 2.39 g Mg � 1 mol Mg

__ 24.31 g Mg

� 0.0983 mol Mg

0.983 mol Mg � 2 mol MgO

__ 2 mol Mg

� 40.31 g MgO

__ 1 mol MgO

� 3.96 g MgO

% yield � 3.48 g MgO

__ 3.96 g MgO

� 100

� 87.9% yield of MgO

Chapter 11 Assessmentpages 392–397

Section 11.1

Mastering Concepts 36. Why must a chemical equation be balanced

before you can determine mole ratios?

Mole ratios are determined by the coefficients in a balanced equation. If the equation is not balanced, the relationship between reactants and products cannot be determined.

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

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

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

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

moles B _ moles A

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

The coefficients in the balanced chemical equation show the numbers of representative particles involved in a reaction. Subscripts give the numbers of different kinds of atoms within a molecule or formula unit.

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

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

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42. When heated by a flame, ammonium dichro-mate decomposes, producing nitrogen gas, solid chromium(III) oxide, and water vapor.

(NH4)2Cr2O7 → N2 � Cr2O3 � 4H2O

Write the mole ratios for this reaction that relate ammonium chromate to the products.

1 mol (NH4)2Cr2O7 __

1 mol N2

1 mol (NH4)2Cr2O7 __ 1 mol Cr2O3

1 mol (NH4)2Cr2O7 __

4 mol H2O

1 mol N2 __

1 mol (NH4)2Cr2O7

1 mol Cr2O3 __ 1 mol (NH4)2Cr2O7

4 mol H2O

__ 1 mol (NH4)2Cr2O7

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

2M2N → M4 � N2

2 mol M2N

__ 1 mol M4

, 2 mol M2N

__ 1 mol N2

,

1 mol M4 _ 1 mol N2

, 1 mol M4 __

2 mol M2N ,

1 mol N2 __

2 mol M2N ,

1 mol N2 _ 1 mol M4

Mastering Problems 44. Interpret the following equation in terms of

particles, moles, and mass.

4Al(s) � 3O2(g) → 2Al2O3(s)

Particles: 4 atoms Al � 3 molecules O2 → 2 formula units Al2O3

Moles: 4 mol Al � 3 mol O2 → 2 mol Al2O3;

Mass:

4Al � 4 mol Al(26.982 g/mol) � 107.93 g

3O2 � 6 mol O(15.999 g/mol) � 95.99 g

2Al2O3 � 4 mol Al(26.982 g/mol) �6 mol O(15.999 g/mol) � 203.92 g

107.93 g Al � 95.99 g O2 → 203.92 g Al2O3

45. Smelting When tin(IV) oxide is heated with carbon in a process called smelting, the element tin can be extracted.

SnO2(s) � 2C(s) → Sn(l) � 2CO(g)

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

Particles: 1 formula unit SnO2 �2 atoms C → 1 atom Sn � 2 molecules CO

Moles: 1 mol SnO2 � 2 molC → 1 mol Sn � 2 mol CO

Mass:

SnO2: 1 mol(118.710 g/mol) �2 mol (15.999 g/mol) � 150.71 g

2C: 2 mol(12.011 g/mol) � 24.02 g

Sn: 1 mol(118.710 g/mol) � 118.71 g

2CO: 2 mol(12.011 g/mol) � 2 mol(15.999 g/mol) � 56.02 g

150.71 g SnO2 � 24.02 g C →118.71 g Sn � 56.02 g CO

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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.

Cu(s) � 4HNO3(aq) → Cu(NO3)2(aq) � 2NO2(g) � 2H2O(l); answers should include any six of the following ratios: 1 mol Cu/4 mol HNO3, 1 mol Cu/1 mol Cu(NO3)2, 1 mol Cu/2 mol NO2, 1 mol Cu/2 mol H2O, 4 mol HNO3/1 mol Cu, 4 mol HNO3/1 mol Cu(NO3)2, 4 mol HNO3/2 mol NO2, 4 mol HNO3/2 mol H2O, 1 mol Cu(NO3)2/1 mol Cu, 1 mol Cu(NO3)2/4 mol HNO3, 1 mol Cu(NO3)2/2 mol NO2, 1 mol Cu(NO3)2/2 mol H2O, 2 mol NO2/1 mol Cu, 2 mol NO2/4 mol HNO3, 2 mol NO2/1 mol Cu(NO3)2, 2 mol NO2/2 mol H2O, 2 mol H2O/1 mol Cu, 2 mol H2O/4 mol HNO3, 2 mol H2O/1 mol Cu(NO3)2, 2 mol H2O/2 mol NO2

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


2HCl(aq) � Pb(NO3)2(aq) → PbCl2(s) � 2HNO3(aq)

b. Interpret the equation in terms of molecules and formula units, moles, and mass.

Particles: 2 molecules HCl � 1 formula unit Pb(NO3)2 → 1 formula unit PbCl2 � 2 molecules HNO3

Moles: 2 mol HCl � 1 mol Pb(NO3)2 → 1 mol PbCl2 � 2 mol HNO3;

Mass:

2HCl: 2 mol(1.008 g/mol) � 2 mol (35.453 g/mol) � 72.9 g

Pb(NO3)2: 1 mol(207.2 g/mol) � 2 mol (14.007 g/mol) � 6 mol(15.999 g/mol) � 331.2 g

PbCl2: 1 mol(207.2 g/mol) � 2 mol(35.453 g/mol) � 278.1 g

2HNO3: 2 mol(1.008 g/mol) � 2 mol(14.007 g/mol) � 6 mol(15.999 g/mol) � 126.0 g

72.9 g HCl � 331.2 g Pb(NO3)2 → 278.1 g PbCl2 � 126.0 g HNO3

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 Fe2O3 are known?

Fe2O3(s) � 2Al(s) → 2Fe(s) � Al2O3(s) � heat

2 mol Fe __ 1 mol Fe2O3

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


SiO2(s) � 4HF(aq) → SiF4(g) � 2H2O(l)

b. List three mole ratios, and explain how you would use them in stoichiometric calculations.

Students may write any of 12 ratios. Examples might be the following. 4 mol HF/1 mol SiO2; used to find the amount of HF that will react with a known amount of SiO2; 1 mol SiF4/1 mol SiO2; used to find the amount of SiF4 that can be formed from a known amount of SiO2. 2 mol H2O/1 mol SiF4; used to find the amount of H2O that will be produced with the SiF4

50. Chrome The most important commercial ore of chromium is chromite (FeCr2O4). 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 (FeCr2).

2C(s) � FeCr2O4(s) → FeCr2(s) � 2CO2(g)

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

1 mol FeCr2 __

1 mol FeCr2O4

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51. Air Pollution The pollutant SO2 is removed from the air 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 SO2 to moles of CaSO4.

2SO2 � 2CaCO3 � O2 → 2CaSO4 � 2CO2

2 mol CaSO4 __

2 mol SO2

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 balance the equation W � X → Y � Z.

Reaction Data

Moles of Reactants Moles of Products

W X Y Z

0.90 0.30 0.60 1.20

Divide each molar quantity by 0.30 mol, the least common denominator:

W: 0.90/0.30 � 3

X: 0.30/0.30 � 1

Y: 0.60/0.30 � 2

Z: 1.20/0.30 � 4

3W � X 0 2Y � 4Z

53. Antacids Magnesium hydroxide is an ingre-dient in some antacids. Antacids react with excess hydrochloric acid in the stomach to relieve indigestion.

___Mg(OH)2 � ___HCl →___ MgCl2 � ___ H2O

a. Balance the reaction of Mg(OH)2 with HCl.

1Mg(OH)2 � 2HCl → 1MgCl2 � 2H2O

b. Write the mole ratio that would be used to determine the number of moles of MgCl2 produced when HCl reacts with Mg(OH)2.

1 mol MgCl2 __

1 mol Mg(OH)2 or

1 mol MgCl2 __ 2 mol HCl

Section 11.2

Mastering Concepts 54. What is the first step in all stoichiometric

calculations?

Write a balanced chemical equation for the reaction.

55. What information does a balanced equation provide?

The balanced equation provides the relationship between reactants and products, and the coefficients in the equation are used to write mole ratios relating reactants and products.

56. On what law is stoichiometry based, and how do the calculations support this law?

Stoichiometry is based on the law of conservation 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. How is molar mass used in some stoichiometric calculations?

Molar mass is a conversion factor for converting moles of a given substance to mass or mass of a given substance to moles.

58. What information must you have in order to calculate the mass of product formed in a chemical reaction?

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.

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+

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 represent hydrogen.


2H2S(g) � O2(g) → 2H2O(g) � 2S(s)

b. Using the same color code, sketch a representation of the flask after the reaction occurs.

Student sketches should show the formation of six water molecules (H2O) and six sulfur atoms (S).

Mastering Problems 60. Ethanol (C2H5OH ), also known as grain

alcohol, can be made from the fermentation of sugar (C6H12O6). The unbalanced chemical equation for the reaction is shown below.

___C6H12O6 → ___C2H5OH � ___CO2

Balance the chemical equation and determine the mass of C2H5OH produced from 750 g of C6H12O6.

C6H12O6 → 2C2H5OH � 2CO2

750 g C6H12O6(180.16 g/mol) � 4.2 mol C6H12O6

4.2 mol C6H12O6 ( 2 mol C2H5OH __

1 mol C6H12O6 )

� 8.4 mol C2H5OH

8.4 mol C2H5OH ( 46.07 g __

1 mol C2H5OH ) � 390 g C2H5OH

61. Welding If 5.50 mol of calcium carbide (CaC2) reacts with an excess of water, how many moles of acetylene (C2H2), a gas used in welding, will be produced?

CaC2(s) � 2H2O(l) → Ca(OH)2(aq) � C2H2(g)

The mole ratio of CaC2:C2H2 is 1:1; thus, 5.50 mol C2H2 will be produced from 5.50 mol CaC2.

62. Antacid Fizz When an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate (NaHCO3), also called sodium bicarbonate, and citric acid (H3C6H5O7).

3NaHCO3(aq) � H3C6H5O7(aq) → 3CO2(g) � 3H2O(l) � Na3C6H5O7(aq)

How many moles of Na3C6H5O7 can be produced if one tablet containing 0.0119 mol of NaHCO3 is dissolved?

0.0119 mol NaHCO3 � (1 mol Na3C6H5O7/3 mol NaHCO3) � 0.00397 mol

63. Esterification The process in which an organic acid and an alcohol that forms as ester and water is known as esterification. Ethyl butanoate (C3H7COOC2H5), an ester, is formed when the alcohol ethanol (C2H5OH) and butanoic acid (C3H7COOH) are heated in the presence of sulfuric acid.

C2H5OH(l) � C3H7COOH(l) → C3H7COOC2H5(l) � H2O(l)

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

4.50 mol C2H5OH � 1 mol C3H7COOC2H5 __

1 mol C2H5OH

� 116.18 g C3H7COOC2H5 ___

1 mol C3H7COOC2H5

� 523 g C3H7COOC2H5

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64. Greenhouse Gas Carbon dioxide is a green-house gas that is linked to global warming. It is released into the atmosphere through the combustion of octane (C8H18) in gasoline. Write the balanced chemical equation for the combus-tion of octane and calculate the mass of octane needed to release 5.00 mol of CO2.

2C8H18(l) � 25O2(g) 0 16CO2(g) � 18H2O(l)

5.00 mol CO2 � 2 mol C8H18 __ 16 mol CO2

� 0.625 mol C8H18

0.625 mol C8H18 � 114.28 g C8H18 __

1 mol C8H18 � 71.4 g C8H18

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.

K2CrO4(aq) � Pb(NO3)2(aq) 0 PbCrO4(s) � 2KNO3(aq)

b. Starting with 0.250 mol of potassium chromate, determine the mass of lead chromate formed.

0.250 mol K2CrO4 � 1 mol PbCrO4 __ 1 mol K2CrO4

� 323.2 g PbCrO4 __ 1 mol PbCrO4

� 80.8 g PbCrO4

66. Rocket Fuel The exothermic reaction between liquid hydrazine (N2H2) and liquid hydrogen peroxide (H2O2) is used to fuel rockets. The products of this reaction are nitrogen gas and water.

a. Write the balanced chemical equation.

N2H2(l) � H2O2(l) 0 N2(g) � 2H2O(g)

b. How much hydrazine, in grams, is needed to produce 10.0 mol of nitrogen gas?

10.0 mol N2 � 1 mol N2H2 __

1 mol N2 �

30.03 g N2H2 __ 1 mol N2H2

� 3.00 � 102 g N2H2

67. Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine.

CH4(g) � 3Cl2(g) 0 CHCl3(g) � 3HCl(g)

How much CH4 in grams is needed to produce 50.0 grams of CHCl3?

50.0 g CHCl3 � 1 mol CHCl3 __

119.37 g CHCl3 �

1 mol CH4 __ 1 mol CHCl3

� 16.04 g CH4 __ 1 mol CH4

� 6.72 g CH4

68. Oxygen Production The Russian Space Agency uses potassium superoxide (KO2) for the chemical oxygen generators in their space suits.

4KO2 � 2H2O � 4CO2 0 4 KHCO3 � 3O2

Complete Table 11.3.

Oxygen Generation Reaction Data

Mass KO2

Mass H2O

MassCO2

Mass KHCO3

Mass O2

1100 g 140 g 7.0 � 102 g 1600 g 380 g

380 g O2 � (1 mol O2/32.00 g O2) � (4 mol KO2/3 mol O2) � (71.1 g KO2/1 mol KO2) � 1100 g

380 g O2 � (1 mol O2/32.00 g O2) � (2 mol H2O/3 mol O2) � (18.02 g H2O/1 mol H2O) � 140 g

380 g O2 � (1 mol O2/32.00 g O2) � (4 mol CO2/3 mol O2) � (44.01 g CO2/1 mol CO2) � 7.0 � 102 g

380 g O2 � (1 mol O2/32.00 g O2) � (4 mol KHCO3/3 mol O2) � (100.12 g KHCO3/1 mol KHCO3) � 1600 g

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69. Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of CO2 produced from the combustion of 100.0 g of ethanol.

C2H5OH(l) � O2(g) 0 CO2(g) � H2O(g)

C2H5OH(l) � 3O2(g) 0 2CO2(g) � 3H2O(l)

100.0 g C2H5OH � 1 mol C2H5OH

__ 46.08 g C2H5OH

� 2.170 mol C2H5OH

2.170 mol C2H5OH � 2 mol CO2 __

1 mol C2H5OH

� 4.340 mol CO2

4.340 mol CO2 � 44.01 g CO2 __ 1 mol CO2

� 191.0 g CO2 produced

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 solution and water.

a. Write the balanced chemical equation for this reaction.

Pb(s) � PbO2(s) � 2H2SO4(aq) 0 2PbSO4(aq) � 2H2O(l)

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.

25 g Pb � 1 mol Pb __ 207.2 g Pb

� 2 mol PbSO4 __

1 mol Pb

� 303.23 g PbSO4 __

1 mol PbSO4 � 73.2 g PbSO4

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) � O2(g) � 2H2O(l) 0 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.

25.0 g NaCN � 1 mol NaCN __ 49.01 g NaCN

� 4 mol Au __ 8 mol NaCN

� 196.97 g Au

__ 1 mol Au

� 50.2 g Au

b. If the mass of the ore from which the gold was extracted is 150.0 g, what percentage of the ore is gold?

50.2 g Au

__ 150.0 g ore

� 100 � 33.5% gold in ore

72. Film Photographic film contains silver bromide in gelatin. Once exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (Na3Ag(S2O3)2) is produced.

AgBr(s) � 2Na2S2O3(aq) 0 Na3Ag(S2O3)2(aq) � NaBr(aq)

Determine the mass of Na3Ag(S2O3)2 produced if 0.275 g of AgBr is removed.

0.275 g AgBr � 1 mol AgBr

__ 187.77 g AgBr

� 1.46 � 10�3 mol AgBr

1.46 � 10�3 mol AgBr � 1 mol Na3Ag(S2O3)2 __

1 mol AgBr

� 1.46 � 10�3 mol Na3Ag(S2O3)2

1.46 � 10�3 mol Na3Ag(S2O3)2

� 401.12 g Na3Ag(S2O3)2 ___

1 mol Na3Ag(S2O3)2 � 0.587 g Na3Ag(S2O3)2

Section 11.3

Mastering Concepts 73. How is a mole ratio used to find the limiting

reactant?

The actual mole ratio of reactants from the chemical equation is compared to the mole ratio determined from the given quantities.

74. Explain why the statement “the limitingreactant is the reactant with the lowest mass” is incorrect.

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.

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

a. Write the balanced equation for the reaction.

3M2 � N2 0 2M3N

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?

6 moles of element M (in the form of 3 moles of M2) and 6 moles of element N (likewise, 3 moles of N2)

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

2 moles of M3N form with 2 moles of N2 unreacted (4 total moles of element N)

d. Identify the limiting reactant and excess reactant.

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

Mastering Problems 76. The reaction between ethyne (C2H2) and

hydrogen (H2) is illustrated in Figure 11.13. The product is ethane (C2H6). Which is the limiting reactant? Which is the excess reactant? Explain.

Ethyne Hydrogen Ethane Ethyne

+ +→

Hydrogen is limiting; ethyne is the excess reactant. One mol of ethyne is left over.

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) � 2H2O(l) 0 Fe(OH)2(s) � 2Ni(OH)2(aq)

How many mol of Fe(OH)2 are produced when 5.00 mol of Fe and 8.00 mol of NiO(OH) react?

According to the balanced equation, two moles of NiO(OH) react with each mole of Fe. So, 4 mol Fe react with 8.00 mol NiO(OH), leaving 1.00 mol Fe in excess. For each mole of Fe that reacts, one mole of Fe(OH)2(s) is produced. Because 4.0 mol Fe reacts, 4.0 mol Fe(OH)2 is produced.

78. One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride?

CsF(s) � XeF6(s) 0 CsXeF7(s).

10.0 mol XeF6 � 1 mol CsXeF7 __

1 mol XeF6 � 10.0 mol CsXeF7

79. Iron Production Iron is obtained commer-cially by the reaction of hematite (Fe2O3) with carbon monoxide. How many grams of iron are produced if 25.0 mol of hematite react with 30.0 mol of carbon monoxide?

Fe2O3(s) � 3CO(g) 0 2Fe(s) � 3CO2(g)

According to the balanced equation, 1 mole of hematite reacts with 3 moles of carbon monoxide. Since 25.0 mol of hematite would require 75.0 mol CO but only 30.0 mol are available, CO is the limiting reactant.

30.0 mol CO � 2 mol Fe _ 3 mol CO

� 55.85 g Fe

_ 1 mol Fe

� 1120 g Fe

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80. The reaction of chlorine gas with solid phosphorus (P4) produces solid phosphorus pentachloride. When 16.0 g of chlorine reacts with 23.0 g of P4, which reactant is limiting? Which reactant is in excess?

10Cl2(g) � P4(s) 0 4PCl5(s)

16.0 g Cl2 � 1 mol CI2 __

70.90 g CI2 � 0.226 mol CI2

23.0 g P4 � 1 mol P4 __

123.88 g P4 � 0.185 moles

According to the balanced equation, Cl2 reacts with P4 in a ten-to-one ratio.

0.226 mol Cl2 � 1 mol P4 _ 10 mol CI2

� 0.0226 mol P4 needed

Cl2 is limiting reactant; P4 is in excess.

81. Alkaline Battery An alkaline battery produces electrical energy according to this equation.

Zn(s) � 2MnO2(s) � H2O(l) 0 Zn(OH)2(s) � Mn2O3(s)

a. Determine the limiting reactant if 25.0 g of Zn and 30.0 g of MnO2 are used.

25.0 g Zn � 1 mol Zn _ 65.3 g Zn

� 0.380 mol Zn

30.0 g MnO2 � 1 mol MnO2 __

86.92 g MnO2

� 0.345 mol MnO2

According to the balanced equation, MnO2 reacts with Zn in a two-to-one ratio. In the reaction, the ratio is 1:1.1 or 0.345/0.380. MnO2 is the limiting reactant.

b. Determine the mass of Zn(OH)2 produced.

0.345 mol MnO2 � 1 mol Zn(OH)2 __ 2 mol MnO2

� 99.39 g Zn(OH)2 __ 1 mol Zn(OH)2

� 17.1 g Zn(OH)2

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

2Li(s) � Br2(l) 0 2LiBr(s)

a. the limiting reactant.

25.0 g Li � 1 mol Li _ 6.94 g Li

� 3.60 mol Li

25.0 g Br2 � 1 mol Br2 __

159.80 g Br2 � 0.156 mol Br2

Actual ratio of mol Li to mol Br2 is 3.60 mol Li/0.156 Br2 or 23:1. Only 2 mol Li are required for 1 mol Br2. Thus, Br2 is the limiting reactant.

b. the mass of lithium bromide produced.

0.156 mol Br2 � 2 mol LiBr _ 1 mol Br2

� 0.312 mol LiBr

0.312 mol LiBr � 86.84 g LiBr

__ 1 mol LiBr

� 27.1 g LiBr

c. the excess reactant and the excess mass.

Li is in excess.

0.156 mol Br2 � 2 mol Li _ 1 mol Br2

� 0.312 mol Li used

3.60 mol Li � 0.312 mol Li used � 3.29 mol Li remaining

3.29 mol Li � 6.94 g Li

_ 1 mol Li

� 22.8 g Li remaining

Section 11.4

Mastering Concepts 83. What is the difference between actual yield and

theoretical yield?

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

84. How are actual yield and theoretical yield determined?

Actual yield is determined through experimentation. Theoretical yield is calculated from a given reactant or the limiting reactant.

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85. Can the percent yield of a chemical reaction be more than 100%? Explain your answer.

No, you cannot produce more product than the theoretical yield, which is determined from the starting reactants.

86. What relationship is used to determine the percent yield of a chemical reaction?

actual yeild

__ theoretical yield

� 100 � percent yield

87. What experimental information do you need in order to calculate both the theoretical and the percent yield of any chemical reaction?

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

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?

the mass of one substance in the reaction and the actual mass of metal hydroxide produced

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.

Element AElement B

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 AB2 particles, but only three were produced. There are enough remaining unreacted A and B particles to produce one more AB2 particle. The percent yield is 75%.

Mastering Problems 90. Ethanol (C2H5OH) is produced from the

fermentation of sucrose (C12H22O11) in the presence of enzymes.

C12H22O11(aq) � H2O(g) 0 4C2H5OH(l) � 4CO2(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.

theoretical yield:

684 g C12H22O11 � 1 mol C12H22O11 __ 342.23 g C12H22O11

�

4 mol C2H5OH

__ 1 mol C12H22O11

� 46.07 g C2H5OH

__ 1 mol C2H5OH

� 369 g C2H5OH

% yield � actual yield


� 100

� 349 g

_ 369 g

� 100 � 94.6%

91. Lead(II) oxide is obtained by roasting galena, lead(II) sulfide, in air. The unbalanced equation is:

PbS(s) � O2(g) 0 PbO(s) � SO2(g)

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

2PbS � 3O2 0 2PbO � 2SO2

Theoretical yield � 200.0 g PbS � 1 mol PbS __ 239.27 g PbS

� 2 mol PbO __ 2 mol PbS

� 223.19 g PbO

__ 1 mol PbO

� 186.6 g PbO

b. What is the percent yield if 170.0 g of PbO is obtained?

% yield � 170.0 g

_ 186.6 g

� 100 � 91.10%

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92. Upon heating, calcium carbonate (CaCO3) decomposes to produce calcium oxide (CaO) and carbon dioxide (CO2).

a. Determine the theoretical yield of CO2 if 235.0 g of CaCO3 is heated.

CaCO3(s) 0 CaO(s) � CO2(g)

235 g CaCO3 � 1 mol CaCO3 __

100.06 g CaCO3

� 1 mol CO2 __ 1 mol CaCO3

� 43.99 g CO2 __ 1 mol CO2

� 103 g CO2

b. What is the percent yield of CO2 if 97.5 g of CO2 is collected?



� 100

� 97.5 g CO2 __ 103 g CO2

� 100 � 94.7%

93. Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosi-licic acid (H2SiF6).

SiO2(s) � 6HF(aq) 0 H2SiF6(aq) � 2H2O(l)

40.0 g SiO2 and 40.0 g HF react to yield 45.8 g H2SiF6.

a. What is the limiting reactant?

40.0 g SiO2 � 1 mol SiO2 __

60.09 g SiO2 � 0.666 mol SiO2

40.0 g HF � 1 mol HF __ 20.01 g HF

� 2.00 mol HF

Ratio of HF to SiO2 in the balanced equation is 6:1. Actual ratio of HF to SiO2 is

2.00 mol HF __ 0.666 mol SiO2

or 3.00:1. Thus, HF is the

limiting reactant.

b. What is the mass of the excess reactant?

2.00 mol HF � 1 mol SiO2 __ 6 mol HF

� 0.333 mol SiO2

0.666 mol SiO2 available � 0.333 mol SiO2 used� 0.333 mol SiO2 in excess

0.333 mol SiO2 � 60.09 g SiO2 __ 1 mol SiO2

� 20.0 g SiO2 in excess

c. What is the theoretical yield of H2SiF6?

2.00 mol HF � 1 mol H2SiF6 __

6 mol HF

� 0.333 mol H2SiF6

0.333 mol H2SiF6 � 144.11 g H2SiF6 __

1 mol H2SiF6

� 48.0 g H2SiF6

48.0 g H2SiF6 is the theoretical yield.

d. What is the percent yield?

% yield � 45.8 g H2SiF6 __ 48.0 H2SiF6

� 100 � 95.4% 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 (ZrI4). In the second step, ZrI4 is decomposed to produce pure zirconium.

ZrI4(s) 0 Zr(s) � 2I2(g)

Determine the percent yield of zirconium if 45.0 g of ZrI4 is decomposed and 5.00 g of pure Zr is obtained.

theoretical yield � 45.0 g ZrI4 � 1 mol ZrI4 __

598.84 g ZrI4 �

1 mol Zr _ 1 mol ZrI4

� 91.22 g Zr

_ 1 mol Zr

� 6.85 g of Zr



� 100

� 5.00 g Zr

_ 6.85 g Zr

� 100 � 73.0%

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95. Methanol, wood alcohol, is produced when carbon monoxide reacts with hydrogen gas.

CO � 2H2 0 CH3OH

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 for this reaction.

Methanol Reaction Data

CO(g) CH3OH(l)

Mass 8.50 g 9.71 g

Molar mass 28.01 g/mol 32.05 g/mol

Moles 0.303 mol 0.303 mol

8.50 g CO � (1 mol CO/28.01 g CO) � 0.303 mol CO

0.303 mol CO � (1 mol CH3OH/1 mol CO) � 0.303 mol CH3OH

0.304 mol CH3OH � (32.05 g CH3OH/1 mol CH3OH) � 9.71 g CH3OH

Percent yield � (8.52 g/9.71 g) � 100 � 87.7% yield

96. Phosphorus (P4) is commercially prepared by heating a mixture of calcium phosphate (CaSiO3), sand (SiO2), and coke (C) in an electric furnace. The process involves two reactions.

2Ca3(PO4)2(s) � 6SiO2(s) 0 6CaSiO3(l) � P4O10(g)

P4O10(g) � 10C(s) 0 P4(g) � 10CO(g)

The P4O10 produced in the first reaction reacts with an excess of coke (C) in the second reac-tion. Determine the theoretical yield of P4 if 250.0 g of Ca3(PO4)2 and 400.0 g of SiO2 are heated. If the actual yield of P4 is 45.0 g, deter-mine the percent yield of P4.

Step 1 is to determine the excess quantity in equation #1.

400.0 g SiO2 � 1 mol SiO2 __

60.08 g SiO2 � 6.657 mol of SiO2

250.0 g Ca3(PO4)2 � 1 mol Ca3(PO4)2 __

310.17 g Ca3(PO4)2

� 0.8060 mol

According to the balanced equation, Ca3(PO4)2 reacts with SiO2 in a one-to-three ratio. In this reaction, SiO2 is in excess and 0.8060 mol of Ca3(PO4)2 react.

Step 2 is to determine amount of P4O10 produced.

0.8060 mol Ca3(PO4)2 � 1 mol P4O10 __

2 mol Ca3(PO4)2

� 0.4030 mol P4O10

Step 3 is to determine amount of P4 produced in step 2 from 0.4030 mol P4O10.

0.4030 mol P4O10 � 1 mol P4 __

1 mol P4O10 �

123.88 g P4 __ 1 mol P4

� 49.92 g of P4

Step 4 is to determine the percent yield, given actual yield is 49.92 g.



� 100

� 45.0 g P4 _ 49.92 g P4

� 100 � 90.1%

97. Chlorine forms from the reaction of hydro-chloric acid with manganese(IV) oxide. The balanced equation is:

MnO2 � 4HCl 0 MnCl2 � Cl2 � 2H2O

Calculate the theoretical yield and the percent yield of chlorine if 86.0 g of MnO2 and 50.0 g of HCl react. The actual yield of Cl2 is 20.0 g.

Step 1: The balanced equation is given.

MnO2 (s) � 4HCl (aq) 0 MnCl2 (aq) � Cl2 (g) � 2H2O (l)

Step 2 is to determine which reactant is in excess.

86.0 g MnO2 � 1 mol MnO2 __

86.94 g MnO2 � 0.989 mol MnO2

50.0 g HCl � 1 mol HCI __ 36.46 g HCI

� 1.37 mol HCI

According to the balanced equation, MnO2 reacts with HCl in a one-to-four ratio. In this reaction, the ratio is 0.989 mol/1.37 mol or 1:1.38. Therefore MnO2 is in excess, and HCl is the limiting reactant.

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Step 3 is to determine the grams of Cl2 produced.

1.37 mol HCl � 1 mol CI2 _ 4 mol HCI

� 70.90 g CI2 __ 1 mol CI2

� 24.3 g CI2.

Step 4 is to determine percent yield, given that 20.0 g of Cl2 is actually formed.

20.0 g CI2 _ 24.3 g CI2

� 100 � 82.3%

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 NH4NO3 produced if the moles of CuS are known?

(NH4)2S � Cu(NO3)2 0 2NH4NO3 � CuS

2 mol NH4NO3 __

1 mol CuS

99. Fertilizer The compound calcium cyanamide (CaNCN) can be used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures.

CaC2(s) � N2(g) 0 CaNCN(s) � C(s)

What mass of CaNCN can be produced if 7.50 mol of CaC2 reacts with 5.00 mol of N2?

5.00 mol N2 � 1 mol CaNCN __ 1 mol N2

� 80.11 g CaNCN

__ 1 mol CaNCN

� 401 g CaNCN

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?

CuO � H2 0 Cu � H2O

32.0 g CuO � 1 mol CuO __ 79.55 g CuO

� 1 mol Cu __ 1 mol CuO

�

63.55 g Cu

__ 1 mol Cu

� 25.6 g Cu

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

a. Write the balanced chemical equation for the formation of nitrogen dioxide from nitrogen oxide.

2NO(g) � O2(g) 0 2NO2(g)

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

2 mol NO2 __ 2 mol NO

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

2H2O(l) 0 H2(g) � O2(g)

36.0 g H2O � (1 mol H2O/18.02 g H2O) � 2.00 mol

theoretical yield � 2.00 mol H2O � 1 mol H2

__ 2 mol H2O

� 2.02 g H2 _ 1 mol H2

� 4.04 g H2

% yield � 3.80 g H2 _ 4.04 g H2

� 100 � 94.1%

103. Iron reacts with oxygen as shown.

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

4Fe(s) 1 3O2(g) 0 2Fe2O3(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?

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The graph levels off because the oxygen limits the reaction at that point.

35.0 g Fe2O3 � 1 mol Fe2O3 __

159.7 g Fe2O3 �

3 mole O2 __ 2 mole Fe2O3

� 0.329 mol O2

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?

No, percent yields cannot be greater than 100%. High results could mean the product was not completely dry, or it was contaminated.

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.

No, because there is only one reactant.

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

Yes, because there are two reactants. Not enough information is given to identify which is the limiting reactant.

106. Design an Experiment Design an experi-ment that can be used to determine the percent yield of anhydrous copper(II) sulfate when copper(II) sulfate pentahydrate is heated to remove water.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 remass. Record. Determine the mass of the anhydrous copper sulfate. Using the equation CuSO4·5H2O 0 CuSO4 � 5H2O and the initial mass of the copper(II) sulfate pentahydrate, determine the theoretical yield of copper(II)

sulfate. Determine the actual yield of copper(II) sulfate. Divide the actual yield by the theoretical yield and multiply by 100 to determine percent yield of copper(II) sulfate.

107. Apply Concepts When a campfire begins to die down and smolder, the flame can be rekindled if you start to fan it. Explain in terms of stoichiometry why the fire begins to flare up again.

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

108. Apply Concepts Students conducted a lab to investigate limiting and excess reac-tants. The students added different volumes of sodium phosphate solution (Na3PO4) to a beaker. They then added a constant volume of cobalt(II) nitrate solution (Co(NO3)2), stirred the contents, and allowed the beakers to sit overnight. The next day, each beaker had a purple precipitate formed at the bottom. The students decanted the supernatant from each beaker, divided it into two samples, and added one drop of sodium phosphate solution to one sample and one drop of cobalt(II) nitrate solu-tion to the second sample. Their results are shown in Table 11.5.

Reaction Data for Co(NO3)2 and Na3PO4

TrialVolumeNa3PO4

VolumeCo(NO3)2

Reaction with Drop of Na3PO4

Reaction with

Drop of Co(NO3)2

1 5.0 mL 10.0 mL purple precipitate

no reaction

2 10.0 mL 10.0 mL no reaction

purple precipitate


purple precipitate


purple precipitate

a. Write a balanced chemical equation for the reaction.

2Na3PO4 (aq) � 3Co(NO3)2 (aq) 0 Co3(PO4)2 (s) � 6NaNO3 (aq)

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b. Based on the results, identify the limiting reactant and the excess reactant for each trial.

Trial 1: Na3PO4 is limiting, Co(NO3)2 is in excess because the addition of Na3PO4 caused an additional reaction.

Trials 2-4: Co(NO3)2 is limiting, Na3PO4 is in excess because the addition of Co(NO3)2 caused and additional reaction.

Challenge 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 hydrogen, 5.38 g of vanadium metal forms.

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

V: 5.38 g __

50.94 g/mol � 0.106 mol

O: 4.21 g __

15.999 g/mol � 0.263 mol

Divided the smallest molar quantity.

0.106 mol _ 0.106 mol

� 1

0.263 mol _ 0.106 mol

� 2.48 � 2.5

Multiply by the mole ratio by 2.2 (1 mol V: 2.5 mol O) � V2O5

V: 5.38 g __

50.94 g/mol � 0.106 mol

O: 3.38 g __

15.999 g/mol � 0.211 mol

Divided the smallest molar quantity.

0.106 mol _ 0.106 mol

� 1

0.211 mol _ 0.106 mol

� 1.99 � 2

1 mol V: 2 mol O � VO2

b. Write balanced equations for the steps of the reaction.

V2O5 � H2 0 2VO2 � H2O

VO2 � 2H2 0 V � 2H2O

c. Determine the mass of hydrogen needed to complete the steps of this reaction.

9.59 g V2O5 __

181.88 g/mol �

1 mol H2 __ 1 mol V2O5

� 2.016 g H2 __ 1 mol H2

� 0.106 g H2

8.76 g VO2 __ 82.94 g/mol

� 2 mol H2 _ 1 mol VO2

� 2.016 g H2 __

1 mol H2

� 0.426 g H2

Total hydrogen mass � 0.106 g � 0.426 g �0.532 g H2

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 hypoth-esis or a theory? Why? (Chapter 1)

a hypothesis, because it is based only on observation, not on data

111. Write the electron configuration for each of the following atoms. (Chapter 5)

a. fluorine

[He]2s22p5

b. aluminum

[Ne]3s23p1

c. titanium

[Ar]4s23d2

d. radon

[Xe]6s24f145d106p6

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112. Explain why the gaseous nonmetals exist as diatomic molecules, but other gaseous elements exist as single atoms. (Chapter 8)

Diatomic molecules achieve a noble gas electron configuration by forming covalent bonds. The monoatomic gases already have noble gas electron configurations.

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

4K(s) � O2(g) 0 2K2O(s)

114. What is the molecular mass of UF6? What is the molar mass of UF6? (Chapter 10)

352.02 amu, 352.02 g/mol

238.03 � 6(18.998) � 352.02

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

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

They are whole number multiples of one another.

b. What is the empirical formula of butanoic acid?

54.5 g C � (1 mol/12.01 g) � 4.54 mol

9.1 g H � (1 mol/1.01 g) � 9.0 mol

36.4 g O � (1 mol/16.00 g ) � 2.28 mol

4.54 mol C/2.28 mol O � 2

9.0 mol H/2.28 mol O � 4

2.28 mol O/2.28 mol O � 1

The empirical formula is C2H4O.

Additional Assessment 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.

Answers will vary. Common pollutants are SO2, NO, NO2 and O3. Check that stoichiometric calculations account for a decrease in the pollutant.

117. Haber Process The percent yield of ammonia produced 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.

Answers will vary. Make sure the equation, N2(g) � 3H2(g) � 2NH3(g) � 92 kJ, is included. The 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.

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Document-Based QuestionsBombardier Beetles Many insects secrete hydrogen peroxide (H2O2) and hydroquinone C6H4(OH)2. Bombardier beetles take this a step further by mixing these chemicals with a catalyst. 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 chemical reaction that results in the bombardier beetle’s defensive spray.

Data obtained from: Becker, Bob. April 2006. ChemMatters.

24: no.2

H2O2

C6H4O2C6H4(OH)2

H2O O2 Energy

OH

OH

O

O

� � � �

Hydroquinone Benzoquinone

Catalyst

118. If the bombardier beetle stores 100.0 mg of hydroquinone (C6H4(OH)2) along with 50.0 mg of hydrogen peroxide (H2O2), what is the limiting reactant?

Balanced reaction:

2C6H4(OH)2 � 4H2O2 0 2C6H4O2 � 6H2O � O2

100 mg 50 mg ? mg

C6H4(OH)2 � H2O2 0 C6H4O2 � H2O � O2 � energy

100.0 mg C6H4(OH)2 � 1 � 10�3 g __ 1 mg

� 1 mol __ 110.00 g C6H4(OH)2

� 9.08 � 10�4 mol

C6H4(OH)2

50.0 mg H2O2 � 1 � 10�3 g __ 1 mg

� 1 mol __ 34.02 g H2O2

�

1.47 � 10�3 mol H2O2

9.08 � 10�4 mol C6H4(OH)2 ___

1.47 � 10�3 mol H2O2

�

0.618 mol C6H4(OH)2 __

1 mol H2O2 from reaction vs.

2 mol C6H4(OH)2 __

4 mol H2O2 from eqauation

H2O2 is the limiting reactant.

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

1.47 � 10�3 mol H2O2 � 2 mol C6H4(OH)2 __

4 mol H2O2 �

110.12 g C6H4(OH)2 __

1 mol C6H4(OH)2 �

1 mg __

1 � 10�3 g

� 80.9 mg C6H4(OH)2 used

100.0 mg � 80.9 mg � 19.1 mg of C6H4(OH)2 in excess

120. How many milligrams of benzoquinone will be produced?

1.47 � 10�3 mol H2O2 � 2 mol C6H4O2 __

4 mol H2O2 �

108.09 g C6H4O2 __

1 mol C6H4O2 �

1 mg __

1 � 10�3 g

� 79.4 mg C6H4O2 produced

Standardized Test Practicepages 398–399

1. Stoichiometry is based on the law of

a. constant mole ratios.b. Avogadro’s constant.c. conservation of energy.d. conservation of mass.

d

Use the graph below to answer Questions 2–5.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

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2. Pure silver metal can be made using the reac-tion shown below.

Cu(s) � 2AgNO3(aq) 0 2Ag(s) � Cu(NO3)2(aq)

How many grams of copper metal will be needed to use up all of the AgNO3 in Dr. Raitano’s laboratory?

a. 18.70 gb. 37.3 gc. 74.7 gd. 100 g

a

100.0 g AgNO3 � 1 mol AgNO3 __

169.88 g AgNO3 � 1mol Cu __

2 mol AgNO3

� 63.55w g Cu

__ 1 mol Cu

� 18.70 g Cu

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

Na2CO3(aq) � Ca(OH)2(aq) 0 2NaOH(aq) � CaCO3(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 molb. 4.72 molc. 8.097 mold. 9.43 mol

c

Find the limiting reactant.

500.0 g Na2CO3 � 1 mol Na2CO3 __

106.00 g Na2CO3 �

4.717 mol Na2CO3

300.0 g Ca(OH)2 � 1 mol Ca2 (OH)2 __ 74.10 g Ca(OH)2

�

4.049 mol Ca(OH)2

Ca(OH)2 is the limiting reactant.

4.049 mol Ca(OH)2 � 2 mol NaOH __ 1 mol Ca(OH)2

�

8.097 mol NaOH

4. Pure O2 gas can be generated from the decom-position of potassium chlorate (KClO3):

2KClO3(s) 0 2KCl(s) � 3O2(g)

If half of the KClO3 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%b. 32.7%c. 65.6%d. 98.0%

b

1 _ 2 (200.0 g KCIO3) � 100.0 g KCIO3

theoretical yield � 100.0 g KCIO3 � 1 mol KCIO3 __

122.5 g KCIO3

� 3 mol O2 __

2 mol KCIO3 �

32.00 g O2 __ 1 mol O2

� 39.17 g O2

percent � actual yield


� 100 � 12.8 g

_ 39.17 g

� 100 � 32.7%

5. Sodium dihydrogen pyrophosphate (Na2H2P2O7), more commonly known as baking powder, is manufactured by heating NaH2PO4 at high temperatures:

2NaH2PO4(s) 0 Na2H2P2O7(s) � H2O(g)

If 444.0 g of Na2H2P2O7 is needed, how much more NaH2PO4 will Dr. Raitano have to buy to make enough Na2H2P2O7?

a. 0.00 gb. 94.0 gc. 130.0 gd. 480 g

c

444.0 g Na2H2P2O7 � 1 mol Na2H2P2O7

__ 221.94 g Na2H2P2O7 �

2 mol NaH2PO4 __

1 mol Na2H2P2O7 �

119.98 g NaH2PO4 __ 1 mol Na2H2PO4

� 48.0 g NaH2PO4

480.0 g � 350.0 g � 130.0 g

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6. Red mercury(II) oxide decomposes at high temperatures to form mercury metal and oxygen gas:

2HgO(s) 0 2Hg(l) � O2(g)

If 3.55 mol of HgO decomposes to form 1.54 mol of O2 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%

d

theoretical yield O2 � 3.55 mol HgO � 1 mol O2 __

2 mol HgO

� 1.775 mol O2

percent yield � 1.54 mol _ 1.775 mol

� 100 � 86.8%

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. Wb. Xc. Yd. Z

c

8. Elements labeled W have their valence electrons in which sublevel?

a. sb. pc. dd. f

b

9. Dimethyl hydrazine (CH3)2N2H2 ignites sponta-neously upon contact with dinitrogen tetroxide (N2O4):

(CH3)2N2H2(l) � 2N2O4(l) 0 3N2(g) � 4H2O(g) � 2CO2(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 consumed in this reaction, how many moles of nitrogen gas will be released?

Set up a mole ratio: 3 mole N2 __

2 moles N2O4

18 moles N2O4 � 3 moles N2 __

2 moles N2O4

� 27 moles N2

Use the table below to answer questions 10 and 11.

First Ionization Energy of Period 3 Elements

Element Atomic Number1st 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

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10. Plot the data from this data table. Place atomic numbers on the x-axis.

Data should form an approximately linear relationship with a few jagged edges, similar to Figure 6.16 on page 191.

1210 14 16 18

Firs

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ergy

(Kj/m

ol)

200

400

600

800

1000

1200

1400

1600

Atomic Number

Atomic Number vs. First Ionization Energy

11. Summarize the general trend in ionization energy. How does ionization energy relate to the number of valence electrons in an element?

Ionization generally increases as you move across a period or row in the periodic table. Elements in the first few families have only 1 or 2 valence electrons, which are relatively easy to remove since this will result in a complete outer shell. Elements on the right side of the periodic table have very high ionization 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.

12. How much cobalt(III) titanate (Co2TiO4), in moles, is in 7.13 g of the compound?

a. 2.39 � 101 molb. 3.10 � 10�2 molc. 3.22 � 101 mold. 4.17 � 10�2 mol

b

7.13 g Co2TiO4 � 1 mol Co2TiO4 __

229.74 g Co2TiO4 0.0310 mol �

3.10 � 10�2 mol

Use the pictures below to answer Questions 13–17.

a.

b.

c.

d.

e.

13. Hydrogen sulfide displays this molecular shape.

a

14. Molecules with this shape have four shared pairs of electrons and no lone pairs of electrons.

c

15. This shape of molecule is referred to as trigonal planar.

b

16. Carbon dioxide displays this molecular shape.

d

17. Molecules with this shape undergo sp2 hybridization.

b

stoichiometrystoichiometry · 2 0 2 formula units mgo ... particles (atoms, molecules, formula...

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