Stoichiometry. Stoichiometry. Section 11.1 Defining SOLUTIONS MANUAL. Practice Problems pages CHAPTER 11. (g) 2MgO(s) (g) 2NH 3.

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Stoichiometry

Section 11.1 Defining

Stoichiometry

pages 368–372

Practice Problems

pages 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.

N

₂

(g)

3H

₂

(g)

→

2NH

₃

(g)

1 molecule N₂ 3 molecules H₂_→ 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 H₂0 34.062 g NH₃ 34.062 g reactants 34.062 g products

b.

HCl(aq)

KOH(aq)

→

KCl(aq)

H

₂

O(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 92.566 g reactants 92.566 g products

c.

2Mg(s)

O

₂

(g)

→

2MgO(s)

2 atoms Mg 1 molecule O₂₀ 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 80.608 g reactants 80.608 g products

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)

___H

₂

O(l)

→

___NaOH(aq)

___H

₂

(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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b.

___Zn(s)

___HNO

₃

(aq)

→

___Zn(NO

₃

)

₂

(aq)

___N

₂

O(g)

___H

₂

O(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 891.68 g reactants 891.68 g products

3. Determine all possible mole ratios for the

following balanced chemical equations.

a.

4Al(s)

3O

₂

(g)

→

2Al

₂

O

₃

(s)

_

4 mol Al 3 mol O₂ 3 mol O₂

__

_{2 mol Al} 2O3

__

2 mol Al2O3 4 mol Al 3 mol O

_

2 4 mol Al 2 mol Al₂O₃

__

3 mol O₂ 4 mol Al

__

2 mol Al₂O₃

b.

3Fe(s)

4H

₂

O(l)

→

Fe

₃

O

₄

(s)

4H

₂

(g)

__

3 mol Fe 4 mol H₂O 3 mol Fe

_

_{4 mol H} 2

__

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 H} 2O 4 mol H

__

2O 4 mol H₂

__

4 mol H2 1 mol Fe₃O₄ 4 mol H₂O

__

_{1 mol Fe} 3O4

__

4 mol H2 4 mol H₂O

c.

2HgO(s)

→

2Hg(l)

O

₂

(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 O} 2

__

2 mol HgO 1 mol O₂

4. Challenge

Balance the following equations,

and determine the possible mole ratios.

a.

ZnO(s)

HCl(aq)

→

ZnCl

₂

(aq)

H

₂

O(l)

ZnO(s) 2HCl(aq) → ZnCl₂(aq) H₂O(l)

__

1 mol ZnO 2 mol HCl 1 mol ZnO

__

_{1 mol ZnCl} 2

__

1 mol ZnO 1 mol H₂O

__

2 mol HCl 1 mol ZnO 2 mol HCl

__

_{1 mol ZnCl} 2

__

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 (C

₄

H

₁₀

)

oxygen

→

carbon dioxide

water

2C₄H₁₀(g) 13O₂(g) ₀ 8CO₂(g) 10H₂O(l)

__

2 mol C4H10 13 mol O₂ 2 mol C₄H₁₀

__

_{8 mol CO} 2

__

2 mol C4H10 10 mol H₂O

__

13 mol O2 2 mol C₄H₁₀ 8 mol CO₂

__

_{2 mol C} 4H10

__

10 mol H2O 2 mol C₄H₁₀

__

10 mol H2O 13 mol O₂ 10 mol H₂O

__

_{8 mol CO} 2

_

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 Assessment

page 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 balanced

chemical equation can be interpreted.

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

8. Apply

The general form of a chemical

reac-tion is

x

A

y

B

→

z

AB. 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/xAand yB/zAB zAB/xAand zAB/yB

9. Apply

Hydrogen peroxide (H

₂

O

₂

) 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, 2H

₂

(g)

O

₂

(g)

→

2H

₂

O. 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

Calculations

pages 373–378

Practice Problems

pages 375–377

11. Methane and sulfur react to produce carbon

disulfide (CS

₂

), a liquid often used in the

production of cellophane.

___CH

₄

(g)

___S

₈

(s)

→

___CS

₂

(l)

___H

₂

S(g)

a.

Balance the equation.

2CH₄(g) S₈(s) _→ 2CS₂(l) 4H₂S(g)

b.

Calculate the moles of CS

₂

produced when

1.50 moles of S

₈

is used.

1.50 mol S₈ 2 mol CS

_

2

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c.

How many moles of H

₂

S are produced?

1.50 mol S₈

_

4 mol H2S

1 mol S₈ 6.00 mol H2S

12. Challenge

Sulfuric acid (H

₂

SO

₄

) is formed

when sulfur dioxide (SO

₂

) 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 H

₂

SO

₄

are produced

from 12.5 moles of SO

₂

?

12.5 mol SO₂

__

2 mol H2SO4

2 mol SO₂ 12.5 mol H₂SO₄ produced

c.

How many moles of O

₂

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 Electric energy

NaCl

Cl

2

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 (TiCl

₄

)

is extracted from titanium oxide (TiO

₂

) using

chlorine and coke (carbon).

TiO

₂

(s)

C(s)

2Cl

₂

(g)

→

TiCl

₄

(s)

CO

₂

(g)

a.

What mass of Cl

₂

gas is needed to react with

1.25 mol of TiO

₂

?

Step 1: Make mole → mole conversion. 1.25 mol TiO₂

__

2 mol Cl2

1 mol TiO₂ 2.50 mol Cl2 Step 2: Make mole _→ mass conversion. 2.50 mol Cl₂ 70.90 g Cl

__

2

1 mol Cl₂ 177 g Cl2

b.

What mass of C is needed to react with

1.25 mol of TiO

₂

?

Step 1: Make mole → mole conversion. 1.25 mol TiO₂

__

1 mol C

1 mol TiO₂ 1.25 mol C Step 2: Make mole → mass conversion. 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 TiO

₂

?

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 inflate

automobile air bags involves sodium azide

(NaN

₃

): 2NaN

₃

(s)

→

2Na(s)

3N

₂

(g).

Determine the mass of N

₂

produced from

the decomposition of NaN

₃

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 Step 2: Make mole _→ mole conversion.

1.538 mol NaN₃

__

3 mol N2

2 mol NaN₃ 2.307 mol N2 Step 3: Make mole → mass conversion. 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 (SO

₂

) reacts with oxygen and water

in the air to form sulfuric acid (H

₂

SO

₄

). Write

the balanced chemical equation for the

reac-tion. If 2.50 g of SO

₂

reacts with excess oxygen

and water, how much H

₂

SO

₄

, 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 Step 3: Make mole → mole conversion. 0.0390 mol SO₂

__

2 mol H2SO4

2 mol SO₂ 0.0390 mol H₂SO₄

Step 4: Make mole → mass conversion. 0.0390 mol H₂SO₄

__

98.09 g H2SO4

1 mol H₂SO₄ 3.83 g H₂SO₄

Section 11.2 Assessment

page 378

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 (Br

₂

) 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:

N

₂

(g)

3H

₂

(g)

→

2NH

₃

(g)

If 2.70 g of H

₂

reacts, how many grams of NH

₃

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 NH

__

3 1 mol NH₃

)

15.2 g NH3

22. Design

a concept map for the following

reaction.

CaCO

₃

(s)

2HCl(aq)

→

CaCl

₂

(aq)

H

₂

O(l)

CO

₂

(g)

The concept map should explain how to

deter-mine the mass of CaCl

₂

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 Reactants

pages 379–384

Practice Problems

page 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)

Fe

₂

O

₃

(s)

→

3Na

₂

O(s)

2Fe(s). If 100.0 g of Na and 100.0 g

of Fe

₂

O

₃

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

Make mole → mole conversion. 0.6261 mol Fe₂O₃

__

6 mol Na

1 mol Fe₂O₃ 3.757 mol Na needed

Make mole → mass conversion. 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 (C

₆

H

₁₂

O

₆

) and oxygen. A plant has

88.0 g of carbon dioxide and 64.0 g of water

available for photosynthesis.

a.

Write the balanced chemical equation for the

reaction.

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b.

Determine the limiting reactant.

Make mass → mole conversion. 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 H} 2O ; 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.

Make mole → mole conversion. 2.00 mol CO₂

__

6 mol H2O

6 mol CO₂ 2.00 mol H2O Make mole → mass conversion.

2.00 mol H₂O 18.02 g H

__

2O 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.

Make mole → mole conversion.

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

Section 11.3 Assessment

page 384

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 (P

₄

S

₃

)

is used in the match heads of some matches. It

is produced in the reaction 8P

₄

3S

₈

→

8P

₄

S

₃

.

Determine which of the following statements

are incorrect, and rewrite the incorrect

state-ments to make them correct.

a.

4 mol P

₄

reacts with 1.5 mol S

₈

to form

4 mol P

₄

S

₃

.

correct

b.

Sulfur is the limiting reactant when 4 mol P

₄

and 4 mol S

₈

react.

Phosphorus is the limiting reactant.

c.

6 mol P

₄

reacts with 6 mol S

₈

, forming

1320 g P

₄

S

₃

.

correct

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Section 11.4 Percent Yield

pages 385–388

Practice Problems

page 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)

→

AlCl

₃

(aq)

3H

₂

O(l). If 14.0 g of

Al(OH)

₃

is present in an antacid tablet,

deter-mine the theoretical yield of AlCl

₃

produced

when the tablet reacts with HCl.

Make mass _→ mole conversion. 14.0 g Al(OH)₃

__

1 mol Al(OH)3

78.0 g Al(OH)₃ 0.179 mol Al(OH)₃

Make mole _→ mole conversion. 0.179 mol Al(OH)₃

__

1 mol AlCl3

1 mol Al(OH)₃ 0.179 mol AlCl₃

Make mole → mass conversion. 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

I

₂

0 ZnI

₂

.

a.

Determine the theoretical yield if 1.912 mol

of zinc is used.

Write the balanced chemical equation. Zn(s) I₂(s) → ZnI₂(s)

Make mole → mole conversion. 1.912 mol Zn 1 mol ZnI

_

2

1 mol Zn 1.912 mol ZnI2 Make mole → mass conversion.

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 (AgNO

₃

), silver crystals

and copper(II) nitrate (Cu(NO

₃

)

₂

) solution

form.

a.

Write the balanced chemical equation for the

reaction.

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.

Make mass _→ mole conversion.

20.0 g Cu

__

1 mol Cu

63.55 g Cu 0.315 mol Cu Make mole → mole conversion. 0.315 mol Cu

_

2 mol Ag

1 mol Cu 0.630 mol Ag Make mole → mass conversion.

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

Section 11.4 Assessment

page 388

31. Identify

which type of yield—theoretical

yield, actual yield, or percent yield—is a

measure of the efficiency of a chemical

reaction.

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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)

→

Fe

₂

S

₃

(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 Fe} 2S3

)

155.9 g Fe₂S₃

35. Calculate

the percent yield of the reaction of

magnesium with excess oxygen:

2Mg(s)

O

₂

(g)

→

2MgO(s)

Reaction Data

Mass of empty crucible 35.67 g Mass of crucible and Mg 38.06 g Mass of crucible and MgO (after

heating)

39.15 g

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 Assessment

pages 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.

(NH

₄

)

₂

Cr

₂

O

₇

→

N

₂

Cr

₂

O

₃

4H

₂

O

Write the mole ratios for this reaction that relate

ammonium chromate to the products.

1 mol (NH

__

4)2Cr2O7 1 mol N₂ 1 mol (NH₄)₂Cr₂O₇

__

1 mol Cr₂O₃ 1 mol (NH

__

4)2Cr2O7 4 mol H₂O

__

1 mol N2 1 mol (NH₄)₂Cr₂O₇ 1 mol Cr₂O₃

__

_{1 mol (NH} 4)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 M

__

2N 1 mol M₄ , 2 mol M₂N

__

1 mol N₂ , 1 mol M

_

4 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)

3O

₂

(g)

→

2Al

₂

O

₃

(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.

SnO

₂

(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 mol

C → 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.

a.

Write the balanced chemical equation for the

reaction.

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 Fe

₂

O

₃

are known?

Fe

₂

O

₃

(s)

2Al(s)

→

2Fe(s)

Al

₂

O

₃

(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.

a.

Write the balanced chemical equation for the

reaction.

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 (FeCr

₂

O

₄

). One of the

steps in the process used to extract chromium

from the ore is the reaction of chromite with

coke (carbon) to produce ferrochrome (FeCr

₂

).

2C(s)

FeCr

₂

O

₄

(s)

→

FeCr

₂

(s)

2CO

₂

(g)

What mole ratio would you use to convert from

moles of chromite to moles of ferrochrome?

__

1 mol FeCr2

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51. Air Pollution

The pollutant SO

₂

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 SO

₂

to moles of CaSO

₄

.

2SO₂ 2CaCO₃ O₂→ 2CaSO₄ 2CO₂

2 mol CaSO

__

4 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

→

___ MgCl

₂

___ H

₂

O

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 MgCl

₂

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.

a.

Write the balanced chemical equation for the

reaction.

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

(C

₂

H

₅

OH ), also known as grain

alcohol, can be made from the fermentation

of sugar (C

₆

H

₁₂

O

₆

). The unbalanced chemical

equation for the reaction is shown below.

___C

₆

H

₁₂

O

₆

→

___C

₂

H

₅

OH

___CO

₂

Balance the chemical equation and determine

the mass of C

₂

H

₅

OH produced from 750 g of

C

₆

H

₁₂

O

₆

.

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 (CaC

₂

)

reacts with an excess of water, how many moles

of acetylene (C

₂

H

₂

), a gas used in welding, will be

produced?

CaC

₂

(s)

2H

₂

O(l)

→

Ca(OH)

₂

(aq)

C

₂

H

₂

(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 (NaHCO

₃

), also

called sodium bicarbonate, and citric acid

(H

₃

C

₆

H

₅

O

₇

).

3NaHCO

₃

(aq)

H

₃

C

₆

H

₅

O

₇

(aq)

→

3CO

₂

(g)

3H

₂

O(l)

Na

₃

C

₆

H

₅

O

₇

(aq)

How many moles of Na

₃

C

₆

H

₅

O

₇

can be

produced if one tablet containing 0.0119 mol of

NaHCO

₃

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 (C

₃

H

₇

COOC

₂

H

₅

), an ester, is formed

when the alcohol ethanol (C

₂

H

₅

OH) and

butanoic acid (C

₃

H

₇

COOH) are heated in the

presence of sulfuric acid.

C

₂

H

₅

OH(l)

C

₃

H

₇

COOH(l)

→

C

₃

H

₇

COOC

₂

H

₅

(l)

H

₂

O(l)

Determine the mass of ethyl butanoate produced

if 4.50 mol of ethanol is used.

4.50 mol C₂H₅OH 1 mol C

__

3H7COOC2H5 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 (C

₈

H

₁₈

) 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 CO

₂

.

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 PbCrO

__

4

1 mol PbCrO₄ 80.8 g PbCrO4

66. Rocket Fuel

The exothermic reaction between

liquid hydrazine (N

₂

H

₂

) and liquid hydrogen

peroxide (H

₂

O

₂

) is used to fuel rockets. The

products of this reaction are nitrogen gas and

water.

a.

Write the balanced chemical equation.

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 N

__

2H2 1 mol N₂ 30.03 g N₂H₂

__

1 mol N₂H₂ 3.00 102_{g N} 2H2

67. Chloroform (CHCl

₃

), an important solvent, is

produced by a reaction between methane and

chlorine.

CH

₄

(g)

3Cl

₂

(g)

0 CHCl

₃

(g)

3HCl(g)

How much CH

₄

in grams is needed to produce

50.0 grams of CHCl

₃

?

50.0 g CHCl₃

__

1 mol CHCl3 119.37 g CHCl₃ 1 mol CH₄

__

_{1 mol CHCl} 3 16.04 g CH

__

4 1 mol CH₄ 6.72 g CH4

68. Oxygen Production

The Russian Space

Agency uses potassium superoxide (KO

₂

) for

the chemical oxygen generators in their space

suits.

4KO

₂

2H

₂

O

4CO

₂

0 4 KHCO

₃

3O

₂

Complete

Table 11.3.

Oxygen Generation Reaction Data

Mass KO₂ Mass H₂O Mass CO₂ 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)

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69. Gasohol

is a mixture of ethanol and gasoline.

Balance the equation, and determine the mass

of CO

₂

produced from the combustion of

100.0 g of ethanol.

C

₂

H

₅

OH(l)

O

₂

(g)

0 CO

₂

(g)

H

₂

O(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)

O

₂

(g)

2H

₂

O(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

(Na

₃

Ag(S

₂

O

₃

)

₂

) is produced.

AgBr(s)

2Na

₂

S

₂

O

₃

(aq) 0

Na

₃

Ag(S

₂

O

₃

)

₂

(aq)

NaBr(aq)

Determine the mass of Na

₃

Ag(S

₂

O

₃

)

₂

produced

if 0.275 g of AgBr is removed.

0.275 g AgBr

__

1 mol AgBr 187.77 g AgBr 1.46 103 mol AgBr

1.46 103 mol AgBr

__

1 mol Na3Ag(S2O3)2 1 mol AgBr 1.46 103 mol Na₃Ag(S₂O₃)₂ 1.46 103 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 limiting

reactant 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 N₂0 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 (C

₂

H

₂

) and

hydrogen (H

₂

) is illustrated in

Figure 11.13.

The product is ethane (C

₂

H

₆

). Which is the

limiting reactant? Which is the excess reactant?

Explain.

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)

2H

₂

O(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 (CsXeF

₇

). How

many moles of CsXeF

₇

can be produced from

the reaction of 12.5 mol of cesium fluoride with

10.0 mol of xenon hexafluoride?

CsF(s)

XeF

₆

(s) 0 CsXeF

₇

(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 (Fe

₂

O

₃

) with

carbon monoxide. How many grams of iron

are produced if 25.0 mol of hematite react with

30.0 mol of carbon monoxide?

Fe

₂

O

₃

(s)

3CO(g)

0 2Fe(s)

3CO

₂

(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