Chapter 9

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 95
CHAPTER 9
CALCULATIONS FROM CHEMICAL EQUATIONS
SOLUTIONS TO REVIEW QUESTIONS
1. A mole ratio is the ratio between the mole amounts of two atoms and/or molecules
involved in a chemical reaction.
2. In order to convert grams to moles the molar mass of the compound under consideration
needs to be determined.
3. The balanced equation is
Ca 3 P 2 + 6 H 2 O ¡ 3 Ca1OH2 2 + 2 PH 3
(a)
Correct:
11 mol Ca 3 P 2 2¢ 2 mol PH 3
1 mol Ca 3 P 2
≤ = 2 mol PH 3
(b) Incorrect: 1 g Ca 3 P 2 would produce 0.4 g PH 3
(c)
(d)
(1 g Ca 3 P 2 )a 1 mol
182.2 g b ¢ 2 mol PH 3
≤ a 33.99 g b = 0.4 g PH
1 mol Ca 3 P 2 mol
3
Correct: see equation
Correct: see equation
(e) Incorrect: 2 mol Ca 3 P 2 requires 12 mol H 2 O to produce 4.0 mol PH 3 .
12 mol Ca 3 P 2 2¢ 6 mol H 2O
1 mol Ca 3 P 2
≤ = 12 mol H 2 O
(f) Correct: 2 mol Ca 3 P 2 will react with 12 mol H 2 O ( 3 mol H 2 O are present in excess)
and 6 mol Ca(OH) 2 will be formed.
12 mol Ca 3 P 2 2¢ 3 mol Ca(OH) 2
1 mol Ca 3 P 2
≤ = 6 mol Ca(OH) 2
(g)
(h)
Incorrect: (200. g Ca 3 P 2 )a 1 mol
182.2 g b ¢ 6 mol H 2O
≤ a 18.02 g b = 119 g H
1 mol Ca 3 P 2 mol
2 O
The amount of water present (100. g) is less than needed to react with
200. g Ca 3 P 2 . H 2 O is the limiting reactant.
Incorrect: water is the limiting reactant.
1100. g H 2 O2a 1 mol
18.02 g b ¢ 2 mol PH 3
6 mol H 2 O ≤ a 33.99 g b = 62.9 g PH
mol
3
- 95 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 96
- Chapter 9 -
4. The balanced equation is
2 CH 4 + 3 O 2 + 2 NH 3 ¡ 2 HCN + 6 H 2 O
(a)
Correct
2 mol HCN
(b) Incorrect: 116 mol O 2 2¢ ≤ = 10.7 mol HCN (not 12 mol HCN)
3 mol O 2
(c)
Correct
(d) Incorrect: 112 mol HCN2¢ 6 mol H 2O
(not 4 mol H 2 O)
2 mol HCN ≤ = 36 mol H 2O
(e) Correct
O 2
(f) Incorrect: is the limiting reactant
2 mol HCN
13 mol O 2 2¢ ≤ = 2 mol HCN
3 mol O 2
(not 3 mol HCN)
5. The theoretical yield of a chemical reaction is the maximum amount of product that can
be produced based on a balanced equation. The actual yield of a reaction is the actual
amount of product obtained.
6. You can calculate the percent yield of a chemical reaction by dividing the actual yield by
the theoretical yield and multiplying by one hundred.
- 96 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 97
CHAPTER 9
SOLUTIONS TO EXERCISES
1. (a)
(b)
(c)
(d)
2. (a)
(b)
(c)
(d)
3. (a)
(b)
(c)
(d)
(e)
125.0 g KNO 3 2a 1 mol
101.1 g b = 0.247 mol KNO 3
1 mol
156 mmol NaOH2a b = 0.056 mol NaOH
1000 mmol
15.4 * 10 2 g (NH 4 ) 2 C 2 O 4 2a 1 mol
124.1 g b = 4.4 mol (NH 4) 2 C 2 O 4
The conversion is: mL sol ¡ g sol ¡ g H 2 SO 4 ¡ mol H 2 SO 4
116.8 mL solution2a 1.727 g
mL b ¢ 0.800 g H 2SO 4
≤ a 1 mol
g solution 98.09 g b = 0.237 mol H 2SO 4
12.10 kg NaHCO 3 2a 1000 g ba 1 mol
kg 84.01 g b = 25.0 mol NaHCO 3
1 g 1 mol
1525 mg ZnCl 2 2a ba
1000 mg 136.3 g b = 3.85 * 10-3 mol ZnCl 2
19.8 * 10 24 1 mol
molecules CO 2 2¢
6.022 * 10 23 molecules ≤ = 16 mol CO 2
1250 mL C 2 H 5 OH2a 0.789 g
mL
12.55 mol Fe(OH) 3 2a 106.9 g b = 273 g Fe(OH) 3
mol
1125 kg CaCO 3 2a 1000 g b = 1.25 * 10 5 g CaCO
kg
3
110.5 mol NH 3 2a 17.03 g b = 179 g NH
mol
3
1 mol
ba
46.07 g b = 4.3 mol C 2H 5 OH
1 mol
172 mmol HCl2a
1000 mmol ba36.46 g b = 2.6 g HCl
mol
1500.0 mL Br 2 2a 3.119 g
mL b = 1559.5 g Br 2 = 1.560 * 10 3 g Br 2
- 97 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 98
- Chapter 9 -
4. (a)
(b)
(c)
(d)
(e)
(0.00844 mol NiSO 4 )a 154.8 g b = 1.31 g NiSO
mol
4
(0.0600 mol HC 2 H 3 O 2 )a 60.05 g b = 3.60 g HC 2 H 3 O 2
mol
(0.725 mol Bi 2 S 3 )a 514.2 g b = 373 g Bi
mol
2 S 3
(4.50 * 10 21 1 mol
molecules C 6 H 12 O 6 )¢
6.022 * 10 23 molecules ≤ a 180.2 g b
mol
= 1.35g C 6 H 12 O 6
(75 mL solution2a 1.175 g
mL ba0.200 g K 2CrO 4
b = 18 g K
g solution
2 CrO 4
5. Larger number of molecules: 10.0 g H 2 O or 10.0 g H 2 O 2
Water has a lower molar mass than hydrogen peroxide. 10.0 grams of water has a lower
molar mass, contains more moles, and therefore more molecules than 10.0 g of H 2 O 2 .
6. Larger number of molecules: 25.0 g HCl or 85 g C 6 H 12 O 6
125.0 g HCl2a 1 mol
36.46 g ba6.022 * 1023 molecules
b = 4.13 * 10 23 molecules HCl
mol
185.0 g C 6 H 12 O 6 2a 1 mol
180.2 g ba6.022 * 1023 molecules
b
mol
HCl contains more molecules
= 2.84 * 10 23 molecules C 6 H 12 O 6
7. Mole ratios
2 C 3 H 7 OH + 9 O 2 ¡ 6 CO 2 + 8 H 2 O
(a)
(b)
(c)
6 mol CO 2
8 mol H 2 O
(d)
2 mol C 3 H 7 OH
2 mol C 3 H 7 OH
2 mol C 3 H 7 OH
6 mol CO 2
(e)
9 mol O 2 8 mol H 2 O
9 mol O 2
8 mol H 2 O
(f)
6 mol CO 2 9 mol O 2
- 98 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 99
- Chapter 9 -
8. Mole ratios
3 CaCl 2 + 2 H 3 PO 4 ¡ Ca 3 (PO 4 ) 2 + 6 HCl
(a)
(b)
(c)
3 mol CaCl 2
1 mol Ca 3 (PO 4 ) 2
(d)
1 mol Ca 3 (PO 4 ) 2
2 mol H 3 PO 4
6 mol HCl
6 mol HCl
(e)
2 mol H 3 PO 4 1 mol Ca 3 (PO 4 ) 2
3 mol CaCl 2
2 mol H 3 PO 4
(f)
2 mol H 3 PO 4 6 mol HCl
9.
C 2 H 5 OH + 3 O 2 ¡ 2 CO 2 + 3 H 2 O
2 mol CO 2
17.75 mol C 2 H 5 OH2¢
1 mol C 2 H 5 OH ≤ = 15.5 mol CO 2
10. The balanced equation is 4 HCl + O 2 ¡ 2 Cl 2 + 2 H 2 O
15.60 mol HCl2¢ 2 mol Cl 2
4 mol HCl ≤ = 2.80 mol Cl 2
11. The balanced equation is
MnO 2 (s) + 4 HCl(aq) ¡ Cl 2 (g) + MnCl 2 (aq) + 2 H 2 O(l)
(a)
(b)
4 mol HCl
11.05 mol MnO 2 2¢ ≤ = 4.20 mol HCl
1 mol MnO 2
11.25 mol H 2 O2¢ 1 mol MnCl 2
2 mol H 2 O ≤ = 0.625 mol MnCl 2s
12.
Al 4 C 3 + 12 H 2 O ¡ 4 Al(OH) 3 + 3 CH 4
(a)
1100. g Al 4 C 3 2a 1 mol
144.0 g b ¢ 12 mol H 2O
1 mol Al 4 C 3
≤ = 8.33 mol H 2 O
(b) 10.600 mol CH 4 2¢ 4 mol Al(OH) 3
3 mol CH 4
≤ = 0.800 mol Al(OH) 3
- 99 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 100
- Chapter 9 -
13. Grams of NaOH
Ca(OH) 2 + Na 2 CO 3 ¡ 2 NaOH + CaCO 3
The conversion is: g Ca(OH) 2 ¡ mol Ca(OH) 2 ¡ mol NaOH ¡ g NaOH
1500 g Ca(OH) 2 2a 1 mol 2 mol NaOH
ba ba 40.00 g b = 5 * 10 2 g NaOH
74.10 g 1 mol Ca(OH) 2 mol
14. Grams of Zn 3 (PO 4 ) 2
3 Zn + 2 H 3 PO 4 ¡ Zn 3 (PO 4 ) 2 + 3 H 2
The conversion is: g Zn ¡ mol Zn ¡ mol Zn 3 (PO 4 ) 2 ¡ g Zn 3 (PO 4 ) 2
110.0 g Zn2a 1 mol
65.39 g b ¢ 1 mol Zn 3(PO 4 ) 2
≤ a 386.1 g b = 19.7 g Zn
3 mol Zn mol
3 (PO 4 ) 2
15. The balanced equation is Fe 2 O 3 + 3 C ¡ 2 Fe + 3 CO
The conversion is: kg Fe 2 O 3 ¡ kmol Fe 2 O 3 ¡ kmol Fe ¡ kg Fe
1125 kg Fe 2 O 3 2a 1 kmol 2 kmol Fe 55.85 kg
ba ba b = 87.4 kg Fe
159.7 kg 1 kmol Fe 2 O 3 kmol
16. The balanced equation is 3 Fe + 4 H 2 O ¡ Fe 3 O 4 + 4 H 2
Calculate the grams of both H 2 O and Fe to produce 375 g Fe 3 O 4
1375 g Fe 3 O 4 2¢ 1 mol
231.6 g ≤ a 4 mol H 2O
≤ a 18.02 g b = 117 g H
1 mol Fe 3 O 4 mol
2 O
1375 g Fe 3 O 4 2a 1 mol 3 mol Fe
b ¢ ≤ a 55.85 g b = 271 g Fe
231.6 g 1 mol Fe 3 O 4 mol
17. The balanced equation is 2 C 2 H 6 + 7 O 2 ¡ 4 CO 2 + 6 H 2 O
(a)
(b)
(c)
115.0 mol C 2 H 6 2¢ 7 mol O 2
2 mol C 2 H 6
≤ = 52.5 mol O 2
18.00 g H 2 O2a 1 mol
18.02 g b ¢ 4 mol CO 2
6 mol H 2 O ≤ a 44.01 g b = 13.0 g CO
mol
2
175.0 g C 2 H 6 2a 1 mol
30.07 g b ¢ 4 mol CO 2
≤ a 44.01 g b = 2.20 * 10 2 g CO
2 mol C 2 H 6 mol
2
(d) 12.75 mol of H 2 O2¢ 4 mol CO 2
6 mol H 2 O ≤ a 44.01 g b = 80.7 g CO
mol
2
- 100 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 101
- Chapter 9 -
(e)
125.0 mol C 2 H 6 2¢ 7 mol O 2
2 mol C 2 H 6
≤ a 32.00 g
mol O 2
b = 2.80 * 10 3 g O 2
(f)
1125 g H 2 O2¢ 1 mol
18.02 g ≤¢2 mol C 2H 6
6 mol H 2 O ≤ a 30.07 g b = 69.5 g C
mol
2 H 6
18.
4 FeS 2 + 11 O 2 ¡ 2 Fe 2 O 3 + 8 SO 2
(a)
11.00 mol FeS 2 2a 2 mol Fe 2O 3
4 mol FeS 2
b = 0.500 mol Fe 2 O 3
(b)
14.50 mol FeS 2 2¢ 11 mol O 2
4 mol FeS 2
≤ = 12.4 mol O 2
(c)
11.55 mol Fe 2 O 3 2¢ 8 mol SO 2
2 mol Fe 2 O 3
≤ = 6.20 mol SO 2
(d)
10.512 mol FeS 2 2¢ 8 mol SO 2
≤ a 64.07 g b = 65.6 g SO
4 mol FeS 2 mol
2
(e)
140.6 g SO 2 2a 1 mol
64.07 g b ¢ 11 mol O 2
8 mol SO 2
≤ = 0.871 mol O 2
(f)
1221 g Fe 2 O 3 2a 1 mol
159.7 g b ¢ 4 mol FeS 2
≤ a 120.0 g b = 332 g FeS
2 mol Fe 2 O 3 mol
2
19. (a) Hydrogen Oxygen
Hydrogen is the limiting reactant.
- 101 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 102
- Chapter 9 -
(b)
Hydrogen Bromine
Bromine is the limiting reactant.
20.
(a)
Lithium Iodine
No limiting reactant.
(b)
Silver Chlorine
Silver is the limiting reactant.
21.
(a)
Potassium Chlorine
Potassium is the limiting reactant.
- 102 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 103
- Chapter 9 -
(b)
Aluminum Oxygen
Oxygen is the limiting reactant.
22.
(a)
Nitrogen Oxygen
Oxygen is the limiting reactant.
(b)
Iron Hydrogen Oxygen
Water is the limiting reactant.
23. (a)
KOH + HNO 3 ¡ KNO 3 + H 2 O
16.0 g 12.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using KNO 3 as the product:
116.0 g KOH2¢ 1 mol
56.10 g ≤ a 1 mol KNO 3
1 mol KOH ≤ a 101.1 g b = 28.8 g KNO
mol
3
112.0 g HNO 3 2a 1 mol
63.02 g b ¢ 1 mol KNO 3
1 mol KOH ≤ a 101.1 g b = 19.3 g KNO
mol
3
Since HNO 3 produces less KNO 3 , it is the limiting reactant and KOH is in excess.
- 103 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 104
- Chapter 9 -
(b)
2 NaOH + H 2 SO 4 ¡ Na 2 SO 4 + 2 H 2 O
10.0 g 10.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2 O as the product:
Since H 2 SO 4 produces less H 2 O, it is the limiting reactant and NaOH is in excess.
24. (a) 2 Bi(NO 3 ) 3 + 3 H 2 S ¡ Bi 2 S 3 + 6 HNO 3
50.0 g 6.00 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using Bi 2 S 3 as the product:
(b)
110.0 g NaOH2a 1 mol
40.00 g b ¢ 2 mol H 2O
2 mol NaOH ≤ a 18.02 g b = 4.51 g H
mol
2 O
110.0 g H 2 SO 4 2a 1 mol
98.09 g b ¢ 2 mol H 2O
≤ a 18.02 g b = 3.67 g H
1 mol H 2 SO 4 mol
2 O
150.0 g Bi(NO 3 ) 3 )a 1 mol
395.0 g ba 1 mol Bi 2S 3
ba 514.2 g b = 32.5 g Bi
2 mol Bi(NO 3 ) 3 mol
2 S 3
(6.00 g H 2 S)a 1 mol
34.09 g b ¢ 1 mol Bi 2S 3
3 mol H 2 S ≤ a 514.2 g b = 30.2 g Bi
mol
2 S 3
Since H 2 S produces less Bi 2 S 3 , it is the limiting reactant and Bi(NO 3 ) 3 is in excess.
3 Fe + 4 H 2 O ¡ Fe 3 O 4 + 4 H 2
40.0 g 16.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2 as the product:
(40.0 g Fe)a 1 mol
55.85 g b ¢ 4 mol H 2
3 mol Fe ≤ a 2.016 g b = 1.93 g H
mol
2
116.0 g H 2 O2a 1 mol
18.02 g b ¢ 4 mol H 2
4 mol H 2 O ≤ a 2.016 g b = 1.79 g H
mol
2
Since H 2 O produces less H 2 , it is the limiting reactant and Fe is in excess.
25. Limiting reactant calculations
C 3 H 8 + 5 O 2 ¡ 3 CO 2 + 4 H 2 O
(a) Reaction between 20.0 g C 3 H 8 and 20.0 g O 2
Convert each amount to moles of CO 2
- 104 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 105
- Chapter 9 -
120.0 g C 3 H 8 2a 1 mol
44.09 g b ¢ 3 mol CO 2
1 mol C 3 H 8
≤ = 1.36 moles CO 2
120.0 g O 2 2a 1 mol
32.00 g b ¢ 3 mol CO 2
5 mol O 2
≤ = 0.375 moles CO 2
O 2 is the limiting reactant. The yield is 0.375 moles CO 2 .
(b) Reaction between 20.0 g C 3 H 8 and 80.0 g O 2
Convert each amount to moles of CO 2
120.0 g C 3 H 8 2a 1 mol
44.09 g b ¢ 3 mol CO 2
1 mol C 3 H 8
≤ = 1.36 moles CO 2
180.0 g O 2 2a 1 mol
32.00 g b ¢ 3 mol CO 2
5 mol O 2
≤ = 1.50 moles CO 2
C 3 H 8 is the limiting reactant. The yield is 1.36 moles CO 2 .
(c) Reaction between 2.0 mol C 3 H 8 and 14.0 mol O 2
According to the equation, 2 mol C 3 H 8 will react with 10 mol O 2 . Therefore, C 3 H 8
is the limiting reactant and 4.0 mol O 2 will remain unreacted.
12.0 mol C 3 H 8 2¢ 3 mol CO 2
1 mol C 3 H 8
≤ = 6.0 mol CO 2 produced
(2.0 mol C 3 H 8 )¢ 4 mol H 2O
1 mol C 3 H 8
≤ = 8.0 mol H 2 O produced
When the reaction is completed, 6.0 mol CO 2 , 8.0 H 2 O, and 4.0 mol O 2 will be in
the container.
26. The balanced equation is 2 C 3 H 6 + 9 O 2 ¡ 6 CO 2 + 6 H 2 O
(a) Reaction between 15.0 g C 3 H 6 and 15.0 g O 2 .
Convert each amount to moles of H 2 O
115.0 g C 3 H 6 2¢ 1 mol
42.08 g ≤¢6 mol H 2O
2 mol C 3 H 6
≤ = 1.07 mol H 2 O
115.0 g O 2 2¢ 1 mol
32.00 g ≤¢6 mol H 2O
2 mol O 2 ≤ = 0.313 mol H 2O
The O 2 is the limiting reactant. The yield is 0.313 mol H 2 O.
- 105 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 106
- Chapter 9 -
(b)
Reaction between 12.0 g of C 3 H 6 and 25.0 g of O 2 .
Convert each amount to moles of H 2 O
112.0 g C 3 H 6 2¢ 1 mol
42.08 g ≤¢6 mol H 2O
2 mol C 3 H 6
≤ = 0.856 mol H 2 O
125.0 g O 2 2¢ 1 mol
32.00 g ≤¢6 mol H 2O
9 mol O 2
≤ = 0.521 mol H 2 O
O 2
is the limiting reactant. The yield is 0.521 mol H 2 O.
(c) Reaction between 5.0 mol of C 3 H 6 and 15.0 mol of O 2 .
Convert each to moles of CO 2
(5.0 mol C 3 H 6 )¢ 6 mol CO 2
2 mol C 3 H 6
≤ = 15 mol CO 2
(15.0 mol O 2 )¢ 6 mol CO 2
≤ = 10 mol CO
9 mol O 2
2
Since is the limiting reactant. C 3 H 6 will be left unreacted.
O 2
27.
28.
X 8 + 12 O 2 ¡ 8 XO 3
The conversion is: g O 2 ¡ mol O 2 ¡ mol X 8
(120.0 g O 2 2a 1 mol
32.00 g b ¢ 1 mol X 8
12 mol O 2
≤ = 0.3125 mol X 8 80.0 g X 8 = 0.3125 mol X 8
80.0 g
0.3125 mol = 256 g>mol X 8
g
256
mol g
molar mass X = = 32.0
8 mol
Using the periodic table we find that the element with 32.0 g>mol is sulfur.
X + 2 HCl ¡ XCl 2 + H 2
The conversion is: g H 2 ¡ mol H 2 ¡ mol X
12.42 g H 2 2a 1 mol
2.016 g b ¢ 1 mol X
1 mol H 2
≤ = 1.20 mol X 78.5 g X = 1.20 molX
78.5 g
= 65.4 g>mol
1.20 mol
Using the periodic table we find that the element with atomic mass 65.4 is zinc.
- 106 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 107
- Chapter 9 -
29. Limiting reactant calculation and percentage yield
2 Al + 3 Br 2 ¡ 2 AlBr 3
Reaction between 25.0 g Al and 100. g Br 2
Calculate the grams of AlBr 3 from each reactant.
125.0 g Al2a 1 mol
26.98 g b ¢ 2 mol AlBr 3
≤ a 266.7 g b = 247 g AlBr
2 mol Al mol
3
(100. g Br 2 )a 1 mol
159.8 g b ¢ 2 mol AlBr 3
≤ a 266.7 g b = 111 g AlBr
3 mol Br 2 mol
3
Br 2 is limiting; 111 g AlBr 3 is the theoretical yield of product.
actual yield
64.2 g
Percent yield = a
b11002 = a b11002 = 57.8%
theoretical yield 111 g
30. Percent yield calculation
Fe(s) + CuSO 4 (aq) ¡ Cu(s) + FeSO 4 (aq)
1400. g CuSO 4 2a 1 mol 1 mol Cu
b ¢ ≤ a 63.55 g b = 159 g Cu (theoretical yield)
159.6 g 1 mol CuSO 4 mol
actual yield
151 g
% yield = a
b11002 = a b11002 = 95.0% yield of Cu
theoretical yield 159 g
31. The balanced equation is 3 C + 2 SO 2 ¡ CS 2 + 2 CO 2
Calculate the g C needed to produce 950 g CS 2 taking into account that the yield of CS 2
is 86.0%. First calculate the theoretical yield of CS 2 .
950 g CS 2
0.860
= 1.1 * 10 3 g CS 2 1theoretical yield2
Now calculate the grams of coke needed to produce 1.1 * 10 3 g CS 2 .
11.1 * 10 3 g CS 2 2a 1 mol
76.15 g b ¢ 3 mol C ≤ a 12.01 g b = 5.2 * 10 2 g C
1 mol CS 2 mol
32. The balanced equation is CaC 2 + 2 H 2 O ¡ C 2 H 2 + Ca1OH2 2
First calculate the grams of pure CaC 2 in the sample from the amount of C 2 H 2 produced.
10.540 mol C 2 H 2 2¢ 1 mol CaC 2
≤ a 64.10 g CaC 2
b = 34.6 g of pure CaC
1 mol C 2 H 2 mol CaC 2 in the sample
2
- 107 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 108
- Chapter 9 -
Now calculate the percent
in the impure sample.
33. No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers,
3600 wrenches and 1200 pliers are needed for 600 tool sets.
34. A subscript is used to indicate the number of atoms in a formula. It cannot be changed
without changing the identity of the substance. Coefficients are used only to balance atoms in
chemical equations. They may be changed as needed to achieve a balanced equation.
35. Consider the reaction A ¡ 2B and assume that you have 1 gram of A. This does not
guarantee that you will produce 1 gram of B because A and B have different molar
masses. One gram of A does not contain the same number of molecules as 1 gram of B.
However, 1 mole of A does have the same number of molecules as one mole of B.
(Remember, 1 mole = 6.022 * 10 23 molecules always.) If you determine the number of
moles in one gram of A and multiply by 2 to get the number of moles of B then from
that you can determine the grams of B using its molar mass. Equations are written in
terms of moles not grams.
36.
(a)
CaC 2
- 108 -
¢ 34.6 g CaC 2
44.5 g sample ≤11002 = 77.8% CaC 2 in the impure sample
4 KO 2 + 2 H 2 O + 4 CO 2 ¡ 4 KHCO 3 + 3 O 2
Á
¢ 0.85 g CO 2
≤ a 1 mol
min 44.01 g b ¢ 4 mol KO 2
≤ = 0.019 mol KO 2
4 mol CO 2 min
¢ 0.019 mol KO 2
b110.0 min2 = 0.19 mol KO
min
2
(b)
37. (a)
The conversion is:
g CO 2
min ¡ mol CO 2
min
¡ mol O 2
min
¢ 0.85 g CO 2
≤ a 1 mol
min 44.01 g b ¢ 3 mol O 2
≤ a 32.00 g
4 mol CO 2 mol
1750 g C 6 H 12 O 6 2a 1 mol
180.2 g b ¢ 2 mol C 2H 5 OH
≤ a 46.07 g b = 380 g C
1 mol C 6 H 12 O 6 mol
2 H 5 OH
ba
¡ g O 2
min ¡ g O 2
hr
60.0 min
b = 28 g O 2
1.0 hr hr
1750 g C 6 H 12 O 6 2a 1 mol
180.2 g b ¢ 2 mol CO 2
≤ a 44.01 g b = 370 g CO
1 mol C 6 H 12 O 6 mol
2
Alternate Solution: 750 g C 2 H 6 O 6 - 380 g
C 2 H 5 OH = 370 g CO 2 by the conservation of mass method.
(b)
1380 g C 2 H 5 OH2a 1 mL
0.79 g b = 480 mL C 2H 5 OH

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 109
- Chapter 9 -
38.
2 CH 3 OH + 3 O 2 ¡ 2 CO 2 + 4 H 2 O
The conversion is:
mL CH 3 OH ¡ g CH 3 OH ¡ mol CH 3 OH ¡ mol O 2 ¡ g O 2
160.0 mL CH 3 OH2a 0.72 g
mL b ¢ 1 mol
32.04 g ≤¢ 3 mol O 2
2 mol CH 3 OH ≤ a 32.00 g b = 65 g O
mol
2
39. The balanced equation is
The conversion is:
7 H 2 O 2 + N 2 H 4 ¡ 2 HNO 3 + 8 H 2 O
(a)
175 kg N 2 H 4 2a 1000 g
1 kg b ¢ 1 mol
32.05 g ≤¢2 mol HNO 3
≤ a 63.02 g b = 2.9 * 10 5 g HNO
1 mol N 2 H 4 mol
3
(b)
(c)
1250 L H 2 O 2 2a
1000 mL
1 L
b ¢ 1.41 g 1 mol
≤¢
1 mL 34.02 g ≤ a 8 mol H 2O
ba 18.02 g
7 mol H 2 O 2 mol
1725 g H 2 O 2 2a 1 mol
34.02 g b ¢ 1 mol N 2H 4
≤ a 32.05 g b = 97.6 g N
7 mol H 2 O 2 mol
2 H 4
b
= 2.1 * 10 5 g H 2 O
(d) Reaction between 750 g of N 2 H 2 and 125 g of H 2 O 2 .
Convert each amount to grams of H 2 O.
1750 g N 2 H 4 2a 1 mol
32.05 g b ¢ 8 mol H 2O
≤ a 18.02 g b = 3.4 * 10 3 g H
1 mol N 2 H 4 mol
2 O
1125 g H 2 O 2 2a 1 mol
34.02 g b ¢ 8 mol H 2O
≤ a 18.02 g b = 75.7 g H
7 mol H 2 O 2 mol
2 O
75.7 g H 2 O can be produced.
(e) Since H 2 O 2 is the limiting reactant, N 2 H 4 is in excess.
1125 g H 2 O 2 2a 1 mol
34.02 g b ¢ 1 mol N 2H 4
≤ a 32.05 g b = 16.8 g N
7 mol H 2 O 2 mol
2 H 4 reacted
750 g N 2 H 4 given - 16.8 g N 2 H 4 used = 730 g N 2 H 4 remaining
- 109 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 110
- Chapter 9 -
40. The balanced equation is
16 HCl + 2 KMnO 4 ¡ 5 Cl 2 + 2 KCl + 2 MnCl 2 + 8 H 2 O
(a) Reaction between 25 g KMnO 4 and 85 g HCl. Convert each to moles of MnCl 2 .
(b)
125 g KMnO 4 2a 1 mol KMnO 4
158.04 g KMnO 4
ba 2 mol MnCl 2
2 mol KMnO 4
b = 0.16 mol MnCl 2
185 g HCl2a 1 mol
36.46 g ba2 mol MnCl 2
16 mol HCl b = 0.29 mol MnCl 2
KMnO 4 is the limiting reactant; 0.16 mol MnCl 2 produced.
175 g KCl2a 1 mol
74.55 g ba8 mol H 2O
2 mol KCl ba18.02 g b = 73 g H
mol
2 O
(c)
Theoretical yield is 91 g Cl 2 ; Percent yield: a 75 g b(100) = 82% yield
91 g
(d) Reaction between 25 g HCl and 25 g KMnO 4 . Convert each amount to grams of CL 2 .
(e)
1150 g HCl2a 1 mol
36.46 g ba 5 mol Cl 2
16 mol HCl ba70.90 g b = 91 g Cl
mol
2
125 g HCl2a 1 mol
36.46 g ba 5 mol Cl 2
16 mol HCl ba70.90 g b = 15 g Cl
mol
2
125 g KMnO 4 2a 1 mol
158.04 g ba 5 mol Cl 2
ba 70.90 g b = 28 g Cl
2 mol KMnO 4 mol
2
HCl is the limiting; KMnO 4 is in excess; 15 g Cl 2 will be produced.
Calculate the mass of unreacted KMnO 4 :
125 g HCl2a 1 mol
.
36.46 g ba2 mol KMnO 4
ba 158.04 g b = 14 g KMnO
16 mol HCl mol
4 will react
Unreacted KMnO 4 = 25 g -14 g = 11 g KMnO 4 remain unreacted.
41. The balanced equation is
4Ag + 2 H 2 S + O 2 ¡ 2 Ag 2 S + 2 H 2 O
(a)
11.1 g Ag2a 1 mol
107.9 g ba2 mol Ag 2S
4 mol Ag ba247.9 g b = 1.3 g Ag
mol
2 S
10.14 g H 2 S2a 1 mol
34.09 g ba2 mol Ag 2S
2 mol H 2 S ba247.9 g b = 1.0 g Ag
mol
2 S
- 110 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 111
- Chapter 9 -
(b)
H 2 S is limiting 1.0 g Ag 2 S forms.
0.17 g - 0.14 g = 0.03 grams more H 2 S needed to completely react Ag.
42. Mass of the beaker
26.500 g beaker + Ca(OH) 2
26.095 g beaker + CaO
0.405 g H 2 O absorbed
10.405 g H 2 O2a 1 mol
18.02 g b = 2.25 * 10-2 mol H 2 O absorbed
Since the reaction is a 1:1 mole, the amount of CaO in the beaker is 2.25 * 10 -2 mol.
Convert to grams.
12.25 * 10 -2 56.08 g CaO
mol CaO2a b = 1.26 g CaO in the beaker.
26.095 g
-1.26 g
24.835 g
10.080 g O 2 2a 1 mol
32.00 g ba2 mol Ag 2S
ba 247.9 g b = 1.2 g Ag
1 mol O 2 mol
2 S
11.1 g Ag2a 1 mol
107.9 g ba2 mol H 2S
4 mol Ag ba34.09 g b = 0.17 g H
mol
2 S reacts
mol
beaker + CaO
CaO
mass of the beaker
43.
Pb(NO 3 ) 2 (aq) + 2 KI(aq) ¡ PbI 2 (s) + 2 KNO 3 (aq)
(a) The solid is lead (II) iodide, PbI 2 .
(b) Double displacement reaction.
(c) Calculate the moles of each reactant.
[15 g Pb(NO 3 ) 2 ] + a 1 mol
331.2 g b = 0.045 mol Pb(NO 3) 2
115 g KI2a 1 mol b = 0.090 mol KI
166.0 g
Stoichiometric quantities of reactants are used.
Theoretical yield of PbI 2 is 0.045 mol.
Actual yield:
Percent yield: a
16.68 g PbI 2 2a 1 mol
461.1 g b = 0.0145 mol PbI 2
0.0145 mol
b11002 = 32% yield
0.045 mol
- 111 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 112
- Chapter 9 -
44. Composition of a mixture of and KCl.
In the mixture only KCl reacts with AgNO 3 .
KCl(aq) + AgNO 3 (aq) ¡ AgCl(s) + KNO 3 (aq)
74.55 g KCl
14.33 g AgCl2a b = 2.25 g KCl in the mixture
143.7 g AgCl
KNO 3
- 112 -
10.00 g mixture - 2.25 g KCl = 7.75 g KNO 3
2.25 g KCl
a b11002 = 22.5% KCl
10.00 g mixture
a 7.75 g KNO 3
10.00 g mixture b11002 = 77.5% KNO 3
45. The balanced equation is Zn + 2 HCl ¡ ZnCl 2 + H 2
180.0 g Zn - 35 g Zn = 145 g Zn reacted with HCl
(a)
(b)
(c)
1145 g Zn2a 1 mol
65.39 g ba1 mol H 2
1 mol Zn b a 2.016 g b = 4.47 g H
mol
2 produced
1145 g Zn2a 1 mol mol HCl
ba2
65.39 g 1 mol Zn ba36.46 g b = 162 g HCl reacted
mol
1180.0 g Zn2a 1 mol mol HCl
ba2
65.39 g 1 mol Zn ba36.46 g b = 201 g HCl reacts
mol
201 g - 162 g = 39 g more HCl needed to react wih the 180.0 g Zn
46.
Fe(s) + CuSO 4 (aq)
2.0 mol 3.0 mol
¡
Cu(s)
FeSO 4 (aq)
(a) 2.0 mol Fe react with 2.0 mol CuSO 4 to yield 2.0 mol Cu and 2.0 mol FeSO 4 .
1.0 mol CuSO 4 is unreacted. At the completion of the reaction, there will be
2.0 mol Cu, 2.0 mol FeSO 4 , and 1.0 mol CuSO 4 .
(b) Determine which reactant is limiting and then calculate the g FeSO 4 produced from
that reactant.
120.0 g Fe2a 1 mol mol Cu
ba1
55.85 g 1 mol Fe ba63.55 g b = 22.8 g Cu
mol
140.0 g CuSO 4 2a 1 mol 1 mol Cu
ba ba 63.55 g b = 15.9 g Cu
159.6 g 1 mol CuSO 4 mol
Since CuSO 4 produces less Cu, it is the limiting reactant. Determine the mass of
FeSO 4 produced from 40.0 g CuSO 4 .
+

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 113
- Chapter 9 -
140.0 g CuSO 4 2a 1 mol
159.6 g b ¢ 1 mol FeSO 4
≤ a 151.9 g b = 38.1 g FeSO
1 mol CuSO 4 mol
4 produced
Calculate the mass of unreacted Fe.
140.0 g CuSO 4 2a 1 mol
159.6 g b ¢ 1 mol Fe
≤ a 55.85 g b = 14.0 g Fe will react
1 mol CuSO 4 mol
Unreacted Fe = 20.0 g - 14.0 g = 6.0 g. Therefore, at the completion of the
reaction, 15.9 g Cu, 38.1 g FeSO 4 , 6.0 g Fe, and no CuSO 4 remain.
47. Limiting reactant calculation
CO(g) + 2 H 2 (g) ¡ CH 3 OH(l)
Reaction between 40.0 g CO and 10.0 g H 2 : determine the limiting reactant by
calculating the amount of CH 3 OH that would be formed from each reactant.
140.0 g CO2a 1 mol
28.01 g b ¢ 1 mol CH 3OH
1 mol CO ≤ a 32.04 g b = 45.8 g CH
mol
3 OH
110.0 g H 2 2a 1 mol
2.016 g b ¢ 1 mol CH 3OH
≤ a 32.04 g b = 79.5 g CH
2 mol H 2 mol
3 OH
CO is limiting; H 2 is in excess; 45.8 g CH 3 OH will be produced.
Calculate the mass of unreacted H 2 :
140.0 g CO2a 1 mol
28.01 g b ¢ 2 mol H 2
1 mol CO ≤ a 2.016 g b = 5.76 g H
mol
2 react
10.0 g H 2 - 5.76 g H 2 = 4.2 g H 2 remain unreacted
48. The balanced equation is C 6 H 12 O 6 ¡ 2 C 2 H 5 OH + 2 CO 2
(a) First calculate the theoretical yield.
1750 g C 6 H 12 O 6 2a 1 mol
180.2 g b ¢ 2 mol C 2H 5 OH
≤ a 46.07 g b
1 mol C 6 H 12 O 6 mol
Then take 84.6% of the theoretical yield to obtain the actual yield.
actual yield =
1theoretical yield2184.62
100
= 3.2 * 10 2 g C 2 H 5 OH
- 113 -
= 3.8 * 10 2 g C 2 H 5 OH (theoretical yield)
= 13.8 * 102 g C 2 H 5 OH2184.62
100

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 114
- Chapter 9 -
(b) 475 g C 2 H 5 OH represents 84.6% of the theoretical yield. Calculate the
theoretical yield.
theoretical yield = 475 g
0.846 = 561 g C 2H 5 OH
Now calculate the g C 6 H 12 O 6 needed to produce 561 g C 2 H 5 OH.
1561 g C 2 H 5 OH2a 1 mol
46.07 g b ¢ 1 mol C 6H 12 O 6
2 mol C 2 H 5 OH ≤ a 180.2 g b = 1.10 * 10 3 g C
mol
6 H 12 O 6
49. The balanced equations are:
CaCl 2 (aq) + 2 AgNO 3 (aq) ¡ Ca(NO 3 ) 2 (aq) + 2 AgCl(s)
MgCl 2 (aq) + 2 AgNO 3 (aq) ¡ Mg1NO 3 2 2 (aq) + 2 AgCl(s)
1 mol of each salt will produce the same amount (2 mol) of AgCl. MgCl 2 has a higher
percentage of Cl than CaCl 2 because Mg has a lower atomic mass than Ca. Therefore, on
an equal mass basis, MgCl 2 will produce more AgCl than will CaCl 2 .
Calculations show that 1.00 g MgCl 2 produces 3.01 g AgCl, and 1.00 g CaCl 2 produces
2.56 g AgCl.
50. The balanced equation is Li 2 O + H 2 O ¡ 2 LiOH
The conversion is: g H 2 O ¡ mol H 2 O ¡ mol Li 2 O ¡ g Li 2 O ¡ kg Li 2 O
¢ 2500 g H 2O
astronaut day ≤ a 1 mol
18.02 g b ¢ 1 mol Li 2O
¢ 4.1 kg Li 2O
astronaut day ≤130 days213 astronauts2 = 3.7 * 102 kg Li 2 O
51. The balanced equation is
H 2 SO 4 + 2 NaCl ¡ Na 2 SO 4 + 2 HCl
First calculate the g HCl to be produced
120.0 L HCl solution2a
Then calculate the g H 2 SO 4 required to produce the HCl
11.01 * 10 4 g HCl2a 1 mol
36.46 g b ¢ 1 mol H 2SO 4
≤ a 98.09 g
2 mol HCl 1 mol b = 1.36 * 104 g H 2 SO 4
Finally, calculate the kg H 2 SO 4 (96%)
1 mol H 2 O ≤ a 29.88 g
mol
ba 1 kg
1000 g b = 4.1 kg Li 2O
astronaut day
1000 mL
ba 1.20 g
1 L 1.00 mL b10.4202 = 1.01 * 104 g HCl
11.36 * 10 4 g H 2 SO 4 2¢ 1.00 g H 2SO 4 solution
0.96 g H 2 SO 4
≤ a 1 kg
1000 g b = 14 kg concentrated H 2SO 4
- 114 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 115
- Chapter 9 -
52. Percent yield of H 2 SO 4
1100.0 g S2a 1 mol b = 3.118 mol S to start with
32.07 g
3.118 mol S ¡ 3.118 mol SO 2 - 10% = 2.806 mol SO 2
-0.3118
2.8062
2.806 mol SO 2 ¡ 2.806 mol SO 3 - 10% = 2.525 mol SO 3
-0.2806
2.5254
2.525 mol SO 3 ¡ 2.525 mol H 2 SO 4 - 10% = 2.273 mol H 2 SO 4
-0.2525
2.2725
12.273 mol H 2 SO 4 2a 98.09 g b = 223.0 g H
mol
2 SO 4 formed
13.118 mol S2a 1 mol H 2SO 4
b = 3.118 mol H
1 mol S
2 SO 4 (theoretical yield)
¢ 2.273 mol H 2SO 4
3.118 mol H 2 SO 4
≤11002 = 72.90% yield
Alternate Solution:
Calculation of yield. There are three chemical steps to the formation of H 2 SO 4 . Each step
has a 10% loss of yield.
Step 1:
Step 2:
Step 3:
100% yield - 10% = 90.00% yield
90.00% yield - 10% = 81.00% yield
81.00% yield - 10% = 72.90% yield
Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 mol
H 2 SO 4 . Therefore 3.118 mol S will give a maximum of 3.118 mol H 2 SO 4 .
13.118 mol S2a 1 mol H 2SO 4
1 mol S
ba 98.09 g b10.72902 = 223.0 g H
mol
2 SO 4 yield
53. According to the equations, the moles of CO 2 come from both reactions and the moles
H 2 O come from only the first reaction.
So the mol NaHCO 3 = 2 * mol H 2 O = 2 * 0.0357 mol = 0.0714 mol NaHCO 3
10.0714 mol NaHCO 3 2a 84.01 g b = 6.00 g NaHCO
mol
3 in the sample
110.00 g NaHCO 3 + Na 2 CO 3 2 - 6.00 g NaHCO 3 = 4.00 g Na 2 CO 3 in the sample
- 115 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 116
- Chapter 9 -
a 6.00 g NaHCO 3
b11002 = 60.0% NaHCO
10.00 g
3
a 4.00 g Na 2CO 3
b11002 = 40.0% Na
10.00 g
2 CO 3
54. The balanced equation is 2 KClO 3 ¡ 2 KCl + 3 O 2
12.82 g mixture - 9.45 g residue = 3.37 g O 2 lost by heating
Because the O 2 lost came only from KClO 3 , we can use it to calculate the amount of
KClO 3 in the mixture.
The conversion is: g O 2 ¡ mol O 2 ¡ mol KClO 3 ¡ g KClO 3
13.37 g O 2 2a 1 mol
32.00 g ba2 mol KClO 3
ba 122.6 g b = 8.61 g KClO
3 mol O 2 mol
3 in the mixture
a 8.61 g KClO 3
12.82 g sample b11002 = 67.2% KClO 3
55. The balanced equation is
Al(OH) 3 (s) + 3 HCl(aq) ¡ AlCl 3 (aq) + 3 H 2 O(l)
The conversion is: L HCl ¡ g HCl ¡ mol HCl ¡ mol Al(OH) 3 : g Al(OH) 3
a 2.5 L
day
g HCl
ba3.0 ba 1 mol
L
36.46 g b ¢ 1 mol Al1OH2 3
3 mol HCl
≤ a 78.00 g b = 5.3 g Al1OH2
mol
3 >day
Now calculate the number of 400. mg tablets that can be made from 5.3 g Al(OH) 3
¢ 5.3 g Al(OH) 3 1000 mg
≤ a ba 1 tablet
day
g 400. mg b = 13 tablets>day
56.
4 P + 5 O 2 ¡ P 4 O 10
P 4 O 10 + 6 H 2 O ¡ 4 H 3 PO 4
In the first reaction:
120.0 g P2a 1 mol b = 0.646 mol P
30.97 g
130.0 g O 2 2a 1 mol
32.00 g b = 0.938 mol O 2
0.646 mol P 3.44 mol P
This is a ratio of
=
0.938 mol O 2 5.00 mol O 2
Therefore, P is the limiting reactant and the P 4 O 10 produced is:
- 116 -

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 117
- Chapter 9 -
10.646 mol P2¢ 1 mol P 4O 10
4 mol P ≤ = 0.162 mol P 4O 10
In the second reaction:
115.0 g H 2 O2a 1 mol
18.02 g b = 0.832 mol H 2O
H 2 O 0.832 mol 5.14 mol
and we have 0.162 mol P 4 O 10 . The ratio of is
=
P 4 O 10 0.162 mol 1.00 mol
Therefore, H 2 O is the limiting reactant and the H 3 PO 4 produced is:
10.832 mol H 2 O2¢ 4 mol H 3PO 4
6 mol H 2 O ≤ a 97.99 g b = 54.4 g H
mol
3 PO 4
- 117 -