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Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
Chapter 4
Stoichiometry: Quantitative Information
about Chemical Reactions
INSTRUCTOR’S NOTES
In this chapter the mathematical concepts of chemical substances are expanded from Chapter 2 to chemical
equations. The chapter also includes stoichiometric calculations for reactions occurring in aqueous solution,
concentration units (molarity), and the pH scale. This book always stresses the notion of the stoichiometric factor in
chemical calculations. We constantly remind our students that everything funnels through this step because reactions
occur in a ratio of the number of atoms or molecules of one substance to the number of atoms or molecules in
another substance, such as 1 to 2 or 3 to 3. But, in the lab, we measure mass, not numbers of moles, so a conversion
between these levels must be done.
Students often have difficulties with limiting reagent problems. For this reason we have incorporated a large number
of solved examples in the text. Correctly calculating the two ratios in Step 2 of the procedure shown in the text is the
key to using the method of this book. The summary table given at the end of each example should help clarify the
concept of a limiting reagent. In class, we present alternate ways of determining the limiting reagent such as
calculating the amount of product that can be produced from each reactant, or the new method introduced in this
edition (page 167), using moles of reaction. We also emphasize percent yield calculations because we find that in
percent yield problems some students reverse the identification of actual and theoretical yields.
Chemical analysis of mixtures and combustion analysis can be challenging. For the latter it may be useful to stress
that the reason for producing CO2 and H2O is to “count” C and H atoms present in the original sample. Of course it
is always a good idea to do as many examples as possible. New to this edition is a section on chemical analysis
using spectrophotometry. This section relates well to the typical laboratory experiments that can be performed at this
point in the course.
Chapter 4 requires approximately four to five lectures.
SUGGESTED DEMONSTRATIONS
1.
Stoichiometry

As an introduction to stoichiometry we have used the decomposition of ammonium nitrate:
NH4NO3(s)  N2O(g) + 2 H2O(g)
This reaction is part of suggested demonstration found in Shakhashiri (Volume 1, page 51). The overall
reaction is spectacular. However, one of the reaction products (ZnO) is irritating, so the reaction must be
done in a very well ventilated room.
65
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions

The reaction of Mg ribbon with O2
2 Mg(s) + O2(g)  2 MgO(s)
is also useful as a demonstration reaction that can be tied to an example of a simple stoichiometry reaction.

A good classroom example for stoichiometry is the decomposition of H 2O2:
2 H2O2()  2 H2O(g) + O2(g)
We use it to power a small “rocket” made as in the following illustration.
A small quantity of MnO2 (1-2 g) is placed in the bottle as a catalyst. (We use a 100-mL bottle and find it
helps if the bottle is damp.) The top is placed in the bottle, and an eye dropper containing 30% H 2O2 is
placed in the open hole of the 2-hole stopper. All of the H2O2 in the dropper is squirted into the bottle. The
decomposition reaction is rapid and exothermic, and the steam and O 2 generated provide the thrust for the
plastic “rocket.” Under the right conditions the “rocket” will fly about 20-30 feet. (CAUTION: Wear rubber
gloves when handling the H2O2 and wash the residue down the drain with a large amount of water.) The
H2O2 rocket has proved to be a very useful lecture demonstration. Not only do the students enjoy it, but it
can be used to illustrate a redox reaction (Chapter 3) and catalysis (Chapter 14).
2.
Limiting Reactants

One useful demonstration to use when discussing a limiting reagent situation is the production of H2 from
Ca in water.
Ca(s) + 2 H2O()  Ca(OH)2(s) + H2(g)
This reaction is a good demonstration because both products are visible: bubbles of H 2 and solid Ca(OH)2.

Add a deficiency, a stoichiometric amount, and an excess of magnesium to three beakers of HCl on an
overhead projector to demonstrate limiting reagents.
66
Chapter 4

Stoichiometry: Quantitative Information about Chemical Reactions
Kashmar, R. J. “The Use of Cut-Out Molecular Models on the Overhead Projector to Illustrate
Stoichiometry and Limiting Reagents,” Journal of Chemical Education 1997, 74, 791.
3.
Combustion Reactions

A very simple demonstration of combustion reactions is the “plastic soda bottle rocket.” Insert large nails
on opposite sides of a plastic soda bottle near the bottom. Place a few milliliters of methanol in the bottle
and shake to saturate the atmosphere in the bottle with alcohol. Place a cork in the top of the bottle (it will
take some experimenting to see how firmly to seat the cork), and then touch one of the nails with a Tesla
coil. The cork will fly out of the bottle with a loud bang.

Fortman, J. J.; Rush, A. C.; Stamper, J. E. “Variations on the ‘Whoosh’ Bottle Alcohol Explosion
Demonstration Including Safety Notes,” Journal of Chemical Education 1999, 76, 1092.
4.
Solution Stoichiometry

One of the biggest problems in teaching stoichiometry is to get the students to connect the words on paper
with actual operations in the laboratory. To help in making this connection, we have included photos in the
text of (a) the preparation of a solution starting with a solid compound (Figure 4.9), preparing a solution by
dilution of a more concentrated one (Figure 4.10), and acid–base and redox titrations (Figure 4.14 and
Example 4.12, respectively). We find it very helpful to demonstrate these processes in class as well.
67
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
SOLUTIONS TO STUDY QUESTIONS
4.1
4.2
4.3
6.0 mol Al ·
3 mol O2
= 4.5 mol O2
4 mol Al
6.0 mol Al ·
2 mol Al2 O3
102 g
·
= 310 g Al2O3
4 mol Al
1 mol Al2 O3
0.750 g Al(OH)3 ·
1 mol Al(OH)3
3 mol HCl
36.46 g
·
·
= 1.05 g HCl
78.00 g
1 mol Al(OH)3 1 mol HCl
0.750 g Al(OH)3 ·
1 mol Al(OH)3
3 mol H2 O
18.02 g
·
·
= 0.520 g H2O
78.00 g
1 mol Al(OH)3 1 mol H2 O
2.56 g Al ·
1 mol Al 3 mol Br2
159.8 g
·
·
= 22.7 g Br2
2 mol Al 1 mol Br2
26.98 g
2.56 g Al + 22.7 g Br2 = 25.3 g Al2Br6
4.4
4.5
(a) 454 g Fe2O3 ·
1 mol Fe2 O3
55.85 g
2 mol Fe
·
·
= 318 g Fe
159.7 g
1 mol Fe2 O3 1 mol Fe
(b) 454 g Fe2O3 ·
28.01 g
1 mol Fe2 O3
3 mol CO
·
·
= 239 g CO
159.7 g
1 mol Fe2 O3 1 mol CO
(a) CO2, carbon dioxide, and H2O, water
(b) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(c) 25.5 g CH4 ·
1 mol CH4
2 mol O2
32.00 g
·
·
= 102 g O2
16.04 g
1 mol CH4 1 mol O2
(d) 25.5 g CH4 + 102 g O2 = 128 g reactants = 128 g products
4.6
(a) BaCl2(aq) + 2 AgNO3(aq)  2 AgCl(s) + Ba(NO3)2(aq)
(b) 0.156 g BaCl2 ·
0.156 g BaCl2 ·
4.7
1 mol BaCl2
169.9 g
2 mol AgNO3
·
·
= 0.255 g AgNO3
208.2 g
1 mol AgNO3
1 mol BaCl2
1 mol BaCl2
143.3 g
2 mol AgCl
·
·
= 0.215 g AgCl
208.2 g
1 mol BaCl2 1 mol AgCl
2.50 mol PbS ·
3 mol O2
= 3.75 mol O2
2 mol PbS
2.50 mol PbS ·
2 mol PbO
= 2.50 mol PbO
2 mol PbS
Equation
Initial amount (mol)
Change (mol)
Amount after reaction (mol)
68
2 PbS(s) +
2.50 mol PbS ·
3 O2(g)  2 PbO(s) + 2 SO2(g)
2.50
3.75
–2.50
–3.75
0
2 mol SO2
= 2.50 mol SO2
2 mol PbS
0
0
0
+2.50
+2.50
2.50
2.50
Chapter 4
4.8
Stoichiometry: Quantitative Information about Chemical Reactions
6.2 mol Fe2O3 ·
3 mol C
= 9.3 mol C
2 mol Fe2 O3
6.2 mol Fe2O3 ·
4 mol Fe
= 12 mol Fe
2 mol Fe2 O3
2 Fe2O3(s) + 3 C(s)  4 Fe(s) +
Equation
Initial amount (mol)
Change (mol)
3 mol CO2
= 9.3 mol CO2
2 mol Fe2 O3
3 CO2(g)
6.2
9.3
0
–6.2
–9.3
+12
+9.3
0
0
12
9.3
Amount after reaction (mol)
4.9
6.2 mol Fe2O3 ·
0
(a) 4 Cr(s) + 3 O2(g)  2 Cr2O3(s)
(b) 0.175 g Cr ·
1 mol Cr 2 mol Cr2 O3
152.0 g
·
·
= 0.256 g Cr2O3
4 mol Cr
52.00 g
1 mol Cr2 O3
(c) 0.256 g Cr2O3 – 0.175 g Cr = 0.081 g O2
Equation
4 Cr(s) + 3 O2(g)
Initial amount (mol)
Change (mol)
Amount after reaction (mol)
4.10
0.00337
0.00253
–0.00337
–0.00253
0

2 Cr2O3(s)
0
+0.00168
0
0.00168
(a) CO2, carbon dioxide, and H2O, water
(b) 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
(c) 13.6 g C2H6 ·
1 mol C2 H6
7 mol O2
32.00 g
·
·
= 50.7 g O2
30.07 g
2 mol C2 H6 1 mol O2
(d) 13.6 g C2H6 + 50.7 g O2 = 64.3 g reactants = 64.3 g products
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
Equation
Initial amount (mol)
Change (mol)
Amount after reaction (mol)
4.11
15 g Na2SO4 ·
0.452
1.58
–0.452
–1.58
0
0
1 mol Na 2SO4
= 0.11 mol Na2SO4
142.0 g
0.63 mol C
5.7 mol C
4 mol C
=
>
0.11 mol Na 2SO4
1 mol Na 2SO4
1 mol Na 2SO4
15 g Na2SO4 ·
0
0
+0.904
+1.36
0.904
1.36
7.5 g C ·
1 mol C
= 0.63 mol C
12.0 g
Na2SO4 is the limiting reactant
1 mol Na 2SO4
1 mol Na 2S
78.0 g
·
·
= 8.2 g Na2S
142.0 g
1 mol Na 2SO4 1 mol Na2 S
69
Chapter 4
4.12
Stoichiometry: Quantitative Information about Chemical Reactions
112 g CaO ·
1 mol CaO
= 2.00 mol CaO
56.08 g
224 g NH4Cl ·
4.19 mol NH 4 Cl
2.10 mol NH4 Cl
2 mol NH4 Cl
=
>
2.00 mol CaO
1 mol CaO
1 mol CaO
112 g CaO ·
CaO is the limiting reactant
1 mol CaO 2 mol NH3
17.03 g
·
·
= 68.0 g NH3
1 mol CaO 1 mol NH3
56.08 g
4.13
35 mol F2
22 mol F2
24 mol F2
=
<
1.6 mol S8
1 mol S8
1 mol S8
4.14
32.0 g S8 ·
(a)
1 mol NH4 Cl
= 4.19 mol NH4Cl
53.49 g
Less F2 is available than required, so F2 is the limiting reactant.
1 mol S8
= 0.125 mol S8
256.5 g
71.0 g Cl2 ·
1.00 mol Cl2
8.00 mol Cl2
4 mol Cl2
=
>
0.125 mol S8
1 mol S8
1 mol S8
1 mol Cl2
= 1.00 mol Cl2
70.91 g
S8 is the limiting reactant
(b) 32.0 g S8 ·
135.0 g
1 mol S8 4 mol S2 Cl2
·
·
= 67.4 g S2Cl2
1 mol S2 Cl2
256.5 g
1mol S8
(c) 32.0 g S8 ·
1 mol S8 4 mol Cl2
70.91g
·
·
= 35.4 g Cl2 required
1 mol Cl2
1mol S8
256.5 g
71.0 g Cl2 available – 35.4 g Cl2 required = 35.6 g Cl2 remains
4.15
(a) 995 g CH4 ·
1 mol CH4
= 62.0 mol CH4
16.04 g
2510 g H2O ·
139 mol H2O
2.24 mol H2O 1 mol H2O
=
>
62.0 mol CH4
1 mol CH4
1 mol CH4
(b) 62.0 mol CH4 ·
1 mol H2 O
= 139 mol H2O
18.02 g
CH4 is the limiting reactant
3 mol H2
2.016 g
·
= 375 g H2
1 mol CH4 1 mol H2
(c) CH4 and H2O react in a 1:1 mole ratio.
139 mol H2O available – 62.0 mol H2O used = 77 mol H2O remains
77 mol H2O ·
70
18.0 g
= 1400 g H2O
1 mol H2 O
Chapter 4
4.16
Stoichiometry: Quantitative Information about Chemical Reactions
(a) 2.70 g Al ·
1 mol Al
= 0.100 mol Al
26.98 g
1 mol Cl2
= 0.0573 mol Cl2
70.91 g
4.06 g Cl2 ·
2 mol Al
0.100 mol Al
1.75 mol Al
>
=
3 mol Cl2
0.0573 mol Cl2
1 mol Cl2
Cl2 is the limiting reactant.
(b) 0.0573 mol Cl2 ·
2 mol AlCl3
133.3 g
·
= 5.09 g AlCl3
3 mol Cl2
1 mol AlCl3
(c) 0.0573 mol Cl2 ·
26.98 g
2 mol Al
·
= 1.03 g Al used
3 mol Cl2 1 mol Al
2.70 g Al available – 1.03 g Al used = 1.67 g Al remains
2 Al(s) + 3 Cl2(g)  2 AlCl3(s)
(d) Equation
Initial amount (mol)
Change (mol)
Amount after reaction (mol)
4.17
0.100
0.0573
–0.0382
–0.0573
0.062
0
0
+0.0382
0.0382
(a) 2 C6H14() + 19 O2(g)  12 CO2(g) + 14 H2O(g)
(b) 215 g C6H14 ·
1 mol C6 H14
= 2.49 mol C6H14
86.18 g
215 g O2 ·
6.72 mol O2
2.70 mol O2
19 mol O2
=
<
2.49 mol C6 H14
1 mol C6 H14
2 mol C6 H14
6.72 mol O2 ·
12 mol CO2
44.01 g
·
= 187 g CO2
1 mol CO2
19 mol O2
6.72 mol O2 ·
14 mol H2 O
18.02 g
·
= 89.2 g H2O
1 mol H2 O
19 mol O2
(c) 6.72 mol O2 ·
1 mol O2
= 6.72 mol O2
32.00 g
O2 is the limiting reactant
2 mol C6 H14
86.18 g
·
= 60.9 g C6H14 used
19 mol O2
1 mol C6 H14
215 g C6H14 available – 60.9 g C6H14 used = 154 g C6H14 remains
2 C6H14() + 19 O2(g)  12 CO2(g) + 14 H2O(g)
(d) Equation
Initial amount (mol)
Change (mol)
Amount after reaction (mol)
4.18
2.49
6.72
–0.707
–6.72
1.78
0
0
0
+4.24
+4.95
4.24
4.95
100. g C7H6O3 ·
1 mol C7 H6 O3 1 mol C9 H8 O4
180.2 g
·
·
= 130. g aspirin
138.1 g
1 mol C7 H6 O3 1 mol C9 H8 O4
100. g C4H6O3 ·
1 mol C4 H6 O3 1 mol C9 H8 O4
180.2 g
·
·
= 176 g aspirin
102.1 g
1 mol C4 H6 O3 1 mol C9 H8 O4
The maximum amount that can be produced is 130. g aspirin.
4.19
332 g
· 100% = 81.6% yield
407 g
71
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
4.20
16.3 g
· 100% = 24.0 % yield
68.0 g
4.21
(a) 10.0 g CuSO4 ·
(b)
4.22
4.23
12.6 g
· 100% = 88.1 % yield
14.3 g
(a) 10.0 g CH3SH ·
(b)
1 mol CuSO4 1 mol Cu(NH3 )4 SO4
227.7 g
·
·
= 14.3 g Cu(NH3)4SO4
1 mol Cu(NH3 )4 SO4
159.6 g
1 mol CuSO4
1 mol CH3SH 1 mol CH3COSCH3
90.15 g
·
·
= 9.37 g CH3COSCH3
48.11 g
2 mol CH3SH
1 mol CH3COSCH3
8.65 g
· 100% = 92.3 % yield
9.37 g
1.245 g mixture – 0.832 g after heating = 0.413 g H2O lost
0.413 g H2O ·
1 mol H2 O 1 mol CuSO4  5 H2O
249.7 g
·
·
= 1.14 g CuSO4·5H2O
18.02 g
5 mol H2O
1 mol CuSO4  5 H2O
1.14 g CuSO4  5 H2 O
· 100% = 91.9 % CuSO4·5H2O
1.245 g mixture
4.24
2.634 g mixture – 2.125 g after heating = 0.509 g H2O lost
0.509 g H2O ·
1 mol H2 O 1 mol CuCl2  2 H2O
170.5 g
·
·
= 2.41 g CuCl2·2 H2O
18.02 g
2 mol H2O
1 mol CuCl2  2 H2O
2.41 g CuCl2  2 H2O
· 100% = 91.4 % CuCl2·2H2O
2.634 g mixture
4.25
0.558 g CO2 ·
1 mol CO2 1 mol CaCO3
100.1 g
·
·
= 1.27 g CaCO3
44.01 g
1 mol CO2
1 mol CaCO3
1.27 g
· 100% = 84.3 % CaCO3
1.506 g
4.26
0.196 g CO2 ·
1 mol CO2
2 mol NaHCO3
84.01 g
·
·
= 0.748 g NaHCO3
44.01 g
1 mol NaHCO3
1 mol CO2
0.748 g
· 100% = 43.5 % NaHCO3
1.7184 g
4.27
0.1964 g TlI ·
1 mol TlI 1 mol Tl2SO4
504.83 g
·
·
= 0.1496 g Tl2SO4
2 mol TlI
331.29 g
1 mol Tl2SO4
0.1496 g
· 100% = 1.467 % Tl2SO4
10.20 g
72
Chapter 4
4.28
Stoichiometry: Quantitative Information about Chemical Reactions
0.127 g Al2O3 ·
1 mol Al2 O3
26.98 g
2 mol Al
·
·
= 0.0672 g Al
102.0 g
1 mol Al2 O3 1 mol Al
0.0672 g
· 100% = 8.79 % Al
0.764 g
4.29
1.481 g CO2 ·
1 mol CO2
1 mol C
·
= 0.03365 mol C
44.010 g
1 mol CO2
0.303 g H2O ·
1 mol H2 O
2 mol H
·
= 0.0336 mol H
18.02 g
1 mol H2 O
0.03365 mol C 1 mol C
=
0.0336 mol H
1 mol H
4.30
0.379 g CO2 ·
The empirical formula is CH
1 mol CO2
1 mol C
·
= 0.00861 mol C
44.01 g
1 mol CO2
0.1035 g H2O ·
1 mol H2 O
2 mol H
·
= 0.01149 mol H
18.015 g
1 mol H2 O
0.01149 mol H 1.33 mol H
4/3 mol H
4 mol H
=
=
=
0.00861 mol C
1 mol C
1 mol C
3 mol C
The empirical formula is C3H4
4.31
(a) 0.300 g CO2 ·
0.123 g H2O ·
1 mol CO2
1 mol C
·
= 0.00682 mol C
44.01 g
1 mol CO2
1 mol H2 O
2 mol H
·
= 0.0137 mol H
18.02 g
1 mol H2 O
0.0137 mol H
2 mol H
=
0.00682 mol C
1 mol C
The empirical formula is CH2
(b)
70.1 g/mol
=5
14.0 g/mol
The molecular formula is C5H10
73
Chapter 4
4.32
Stoichiometry: Quantitative Information about Chemical Reactions
(a) 0.364 g CO2 ·
1 mol CO2
1 mol C
·
= 0.00827 mol C
44.01 g
1 mol CO2
0.0596 g H2O ·
1 mol H2 O
2 mol H
·
= 0.00661 mol H
18.02 g
1 mol H2 O
0.00827 mol C 1.25 mol C
5/4 mol C
5 mol C
=
=
=
0.00661 mol H
1 mol H
1 mol H
4 mol H
The empirical formula is C5H4
(b)
4.33
128.2 g/mol
=2
64.09 g/mol
0.1356 g CO2 ·
The molecular formula is C10H8
1 mol CO2
1 mol C
·
= 0.003081 mol C
44.010 g
1 mol CO2
0.003081 mol C ·
12.011 g
= 0.03701 g C
1 mol C
0.0833 g H2O ·
1 mol H2 O
2 mol H
·
= 0.00925 mol H
18.02 g
1 mol H2 O
0.00925 mol H ·
1.008 g
= 0.00932 g H
1 mol H
mass of O = sample mass – mass of C – mass of H
= 0.0956 g – 0.03701 g C – 0.00932 g H
= 0.0492 g O
0.0492 g O ·
1 mol O
= 0.00307 mol O
15.999 g
0.003081 mol C 1 mol C
=
0.00307 mol O
1 mol O
0.00925 mol H
3 mol H
=
0.00307 mol O
1 mol O
The empirical formula is CH3O
62.1 g/mol
=2
31.0 g/mol
4.34
0.3718 g CO2 ·
The molecular formula is C2H6O2
1 mol CO2
1 mol C
·
= 0.008448 mol C
44.010 g
1 mol CO2
0.008448 mol C ·
74
12.011 g
= 0.1015 g C
1 mol C
0.1522 g H2O ·
1 mol H2 O
2 mol H
·
= 0.01690 mol H
18.015 g
1 mol H2 O
0.01690 mol H ·
1.0079 g
= 0.01703 g H
1 mol H
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
mass of O = sample mass – mass of C – mass of H
= 0.1523 g – 0.1015 g C – 0.01703 g H
= 0.0338 g O
0.0338 g O ·
1 mol O
= 0.00211 mol O
16.00 g
0.008448 mol C
4 mol C
=
0.00211 mol O
1 mol O
0.01690 mol H
8 mol H
=
0.00211 mol O
1 mol O
The empirical formula is C4H8O
The empirical formula mass is equal to the molar mass. The molecular formula is also C 4H8O.
4.35
0.0426 g NiO ·
0.100 g CO2 ·
1 mol Ni
1 mol NiO
·
= 5.70  10–4 mol Ni
1 mol NiO
74.69 g
1 mol CO2
1 mol CO
·
= 2.27  10–3 mol CO
44.01 g
1 mol CO2
2.27  10–3 mol CO 4 mol CO
=
5.70  10–4 mol Ni
1 mol Ni
4.36
0.799 g Fe2O3 ·
2.200 g CO2 ·
The empirical formula is Ni(CO)4
1 mol Fe2 O3
2 mol Fe
·
= 0.0100 mol Fe
159.7 g
1 mol Fe2 O3
1 mol CO2
1 mol CO
·
= 0.04999 mol CO
44.010 g
1 mol CO2
0.04999 mol CO
5 mol CO
=
0.0100 mol Fe
1 mol Fe
4.37
6.73 g Na2CO3 ·
The empirical formula is Fe(CO)5
1 mol Na 2 CO3
= 0.0635 mol Na2CO3
106.0 g
0.0635 mol Na 2 CO3
= 0.254 M Na2CO3
0.250 L
[Na+] = 2  [Na2CO3] = 0.508 M Na+
4.38
2.335 g K2Cr2O7 ·
[CO32–] = [Na2CO3] = 0.254 M CO32–
1 mol K2 Cr2 O7
= 0.007937 mol K2Cr2O7
294.18 g
0.007937 mol K 2 Cr2 O7
= 0.0159 M K2Cr2O7
0.500 L
[K+] = 2  [K2Cr2O7] = 0.0318 M
4.39
0.250 L ·
[Cr2O72–] = [K2Cr2O7] = 0.0159 M
0.0125 mol KMnO4
158.0 g
·
= 0.494 g KMnO4
1L
1 mol KMnO4
75
Chapter 4
4.40
Stoichiometry: Quantitative Information about Chemical Reactions
0.125 L ·
1.023  10–3 mol Na3 PO4
163.9 g
·
= 0.0210 g Na3PO4
1L
1 mol Na3 PO4
[Na+] = 3  [Na3PO4] = 3.069  10–3 M
[PO43–] = [Na3PO4] = 1.023  10–3 M
4.41
25.0 g NaOH ·
103 mL
1L
1 mol NaOH
·
·
= 5080 mL solution
1L
0.123 mol NaOH
40.00 g
4.42
322 g KMnO4 ·
1 mol KMnO4
1L
·
= 0.989 L solution
158.0 g
2.06 mol KMnO4
4.43
(a) 0.50 M NH4+; 0.25 M SO42–
(b) 0.246 M Na+; 0.123 M CO32–
(c) 0.056 M H3O+; 0.056 M NO3–
4.44
(a) 0.12 M Ba2+; 0.24 M Cl–
(b) 0.0125 M Cu2+; 0.0125 M SO42–
(c) 1.000 M K+; 0.500 M Cr2O72–
4.45
500.0 mL ·
0.0200 mol Na 2CO3
1L
105.99 g
·
·
= 1.06 g Na2CO3
103 mL
1L
1 mol Na2 CO3
Weigh out 1.06 g of Na2CO3 and place it in the 500.0 mL flask. Add a small amount of distilled water and
mix until the solute dissolves. Add water until the meniscus of the solution rests at the calibrated mark on
the neck of the volumetric flask. Cap the flask and swirl to ensure adequate mixing.
0.15 mol H2 C2 O4
90.04 g
·
= 3.4 g H2C2O4
1L
1 mol H2 C2 O4
4.46
0.250 L ·
4.47
cd = c c ·
25.0 mL
Vc
= 1.50 M ·
= 0.0750 M HCl
500. mL
Vd
4.48
cd = c c ·
4.00 mL
Vc
= 0.0250 M ·
= 0.0100 M CuSO4
10.0 mL
Vd
4.49
(a) cd = cc ·
0.0208 L
Vc
= 6.00 M ·
= 0.125 M H2SO4
1.00 L
Vd
(b) cd = cc ·
0.0500 L
Vc
= 3.00 M ·
= 0.150 M H2SO4
1.00 L
Vd
(a) cd = cc ·
30.0 mL
Vc
= 1.50 M ·
= 0.150 M K2Cr2O7
300. mL
Vd
(b) cd = cc ·
250. mL
Vc
= 0.600 M ·
= 0.500 M K2Cr2O7
300. mL
Vd
4.50
76
Correct method
Correct method
Chapter 4
4.51
4.52
Stoichiometry: Quantitative Information about Chemical Reactions
First dilution: cd = cc ·
25.00 mL
Vc
= 0.136 M ·
= 0.0340 M HCl
100.00 mL
Vd
Second dilution: cd = cc ·
10.00 mL
Vc
= 0.0340 M ·
= 0.00340 M HCl
100.00 mL
Vd
Second dilution: cc = cd ·
100.00 mL
Vd
= 0.000158 M ·
= 0.00316 M
5.00 mL
Vc
First dilution: cc = cd ·
100.00 mL
Vd
= 0.00316 M ·
= 0.158 M
2.00 mL
Vc
4.53
[H3O+] = 10–pH = 10–3.40 = 4.0  10–4 M
The solution is acidic (pH < 7)
4.54
[H3O+] = 10–pH = 10–10.5 = 3  10–11 M
The solution is basic (pH > 7)
4.55
[H3O+] = [HNO3] = 0.0013 M
pH = –log[H3O+] = –log(0.0013) = 2.89
4.56
[H3O+] = [HClO4] = 1.2  10–4 M
pH = –log[H3O+] = –log(1.2  10–4) = 3.92
4.57
pH
(a)
1.00
[H3O+]
0.10 M
3.2 
10–11
acidic
(b)
10.50
(c)
4.89
1.3  10–5 M
acidic
(d)
7.64
2.3  10–8 M
basic
4.58
pH
M
basic
[H3O+]
(a)
9.17
6.7  10–10 M
basic
(b)
5.66
2.2  10–6 M
acidic
(c)
5.25
5.6  10–6 M
acidic
(d)
1.60
2.5  10–2 M
acidic
103 mL
1 mol Ba(OH)2
2 mol HNO3
1L
·
·
·
= 268 mL solution
1L
171.3 g
1 mol Ba(OH)2 0.109 mol HNO3
4.59
2.50 g Ba(OH)2 ·
4.60
50.0 mL ·
4.61
15.0 L ·
40.00 g
0.35 mol NaCl 2 mol NaOH
·
·
= 210 g NaOH
2 mol NaCl
1 mol NaOH
1L
15.0 L ·
0.35 mol NaCl
1 mol Cl2
70.91 g
·
·
= 190 g Cl2
2 mol NaCl 1 mol Cl2
1L
0.125 mol HNO3 1 mol Na2 CO3
1L
106.0 g
·
·
·
= 0.331 g Na2CO3
3
10 mL
1L
2 mol HNO3
1 mol Na2 CO3
77
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
0.146 mol H2SO4
1L
2 mol N2 H4
32.05 g
·
·
·
= 2.34 g N2H4
3
10 mL
1L
1 mol H2SO4 1 mol N2 H4
4.62
250. mL ·
4.63
0.225 g AgBr ·
4.64
2.05 g Al ·
103 mL
1 mol AgBr 2 mol Na2 S2 O3
1L
·
·
·
= 174 mL solution
1L
187.8 g
1 mol AgBr
0.0138 mol Na2 S2 O3
1 mol Al 2 mol KAl(OH)4
134.1 g
·
·
= 10.2 g KAl(OH)4
2 mol Al
1 mol KAl(OH)4
26.98 g
1.35 mol KOH 2 mol KAl(OH)4
134.1 g
·
·
= 33.5 g KAl(OH)4
2 mol KOH
1 mol KAl(OH)4
1L
0.185 L ·
Al is the limiting reactant, so none will remain. The mass of KAl(OH)4 produced in the reaction is 10.2 g.
103 mL
2.25 mol NaCl 1 mol Pb(NO3 )2
1L
·
·
·
= 1.50  103 mL solution
1L
2 mol NaCl
1L
0.750 mol Pb(NO3 )2
4.65
1.00 L ·
4.66
35.2 mL ·
4.67
1.45 g NaOH ·
4.68
2.152 g Na2CO3 ·
103 mL
1 mol Na 2 CO3
1L
2 mol HCl
·
·
·
= 42.5 mL
1L
105.99 g
1 mol Na2 CO3 0.955 mol HCl
4.69
2.150 g Na2CO3 ·
1
1 mol Na 2 CO3
2 mol HCl
·
·
= 1.052 M HCl
105.99 g
1 mol Na2 CO3 0.03855 L
4.70
0.902 g KHC8H4O4 ·
4.71
36.04 mL ·
0.546 mol NaOH 1 mol H2 C2 O4
1L
1L
·
·
·
= 0.0769 L
3
2 mol NaOH
10 mL
1L
0.125 mol H2 C2 O4
103 mL
1 mol NaOH
1 mol HCl
1L
·
·
·
= 44.6 mL
1L
1 mol NaOH 0.812 mol HCl
40.00 g
1
1 mol KHC8 H4 O4
1 mol NaOH
·
·
= 0.167 M NaOH
204.22 g
1 mol KHC8 H4 O4 0.02645 L
0.509 mol NaOH
1 mol H2 A
1L
·
·
= 0.00917 mol H2A
2 mol NaOH
103 mL
1L
0.954 g
= 104 g/mol
0.00917 mol
4.72
29.1 mL ·
0.513 mol NaOH
1L
·
= 0.0149 mol NaOH
103 mL
1L
0.0149 mol NaOH ·
1 mol citric acid
= 0.00498 mol acid
3 mol NaOH
0.956 g
= 192 g/mol
0.00498 mol
0.0149 mol NaOH ·
1 mol tartaric acid
= 0.00746 mol acid
2 mol NaOH
0.956 g
= 128 g/mol
0.00746 mol
The calculated molar mass matches that of citric acid (192 g/mol) but not that of tartaric acid (150. g/mol),
so the unknown acid is citric acid.
78
Chapter 4
4.73
Stoichiometry: Quantitative Information about Chemical Reactions
22.25 mL ·
0.0123 mol KMnO4
55.85 g
1L
5 mol Fe2+
·
·
= 0.0764 g Fe
3
10 mL
1L
1 mol KMnO4 1 mol Fe2+
0.0764 g Fe
· 100% = 12.8% Fe
0.598 g sample
27.85 mL ·
4.75
(a)
Absorbance
4.74
0.102 mol Br2 1 mol C6 H8 O6
1L
176.13 g
·
·
·
= 0.500 g C6H8O6
103 mL
1L
1 mol Br2
1 mol C6 H8 O6
0.8
0.6
0.4
0.2
y = 115000x + 0.1785
0
0
0.000002 0.000004 0.000006
dye concentration (mol/L)
Slope is 1.2 x 105 with an intercept of 0.18.
(b) 0.52 = 1.15  105(dye concentration) + 0.1785
dye concentration = 3.0  10–6 M
(a)
Absorbance
4.76
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
y = 33375x + 0.0023
0
5E-06 1E-05 2E-05 2E-05
nitrite concentration (mol/L)
(b) 0.402 = 3.3375  104(nitrite ion concentration) + 0.0023
nitrite ion concentration = 1.20  10–5 M
4.77
(a) CO2, carbon dioxide, and H2O, water
(b) 2 C6H6() + 15 O2(g)  12 CO2(g) + 6 H2O()
(c) 16.04 g C6H6 ·
1 mol C6 H6
15 mol O2
31.999 g
·
·
= 49.28 g O2
78.113 g
2 mol C6 H6 1 mol O2
(d) 16.04 g C6H6 + 49.28 g O2 = 65.32 g products
79
Chapter 4
4.78
Stoichiometry: Quantitative Information about Chemical Reactions
125 mg acetoacetic acid ·
1g
1 mol acetoacetic acid
1 mol acetone
58.08 g
·
·
·
3
1 mol acetoacetic acid 1 mol acetone
10 mg
102.1 g
= 0.0711 g acetone
4.79
4.80
95 mg urea ·
1g
1 mol urea 1 mol arginine
174 g
·
·
·
= 0.28 g arginine
1 mol urea
60.1 g
1 mol arginine
103 mg
95 mg urea ·
1g
1 mol urea 1 mol ornithine
132 g
·
·
·
= 0.21 g ornithine
1 mol urea
1 mol ornithine
60.1 g
103 mg
(a) 2 Fe(s) + 3 Cl2(g)  2 FeCl3(s)
(b) 10.0 g Fe ·
10.0 g Fe ·
(c)
1 mol Fe 3 mol Cl2
70.91 g
·
·
= 19.0 g Cl2
2 mol Fe
55.85 g
1 mol Cl2
1 mol Fe 2 mol FeCl3
162.2 g
·
·
= 29.0 g FeCl3
2 mol Fe
55.85 g
1 mol FeCl3
18.5 g
· 100% = 63.7% yield
29.0 g
(d) 10.0 g Fe requires 19.0 g Cl2 for complete reaction, so chlorine is the limiting reactant if 10.0 g of
each reactant is combined.
10.0 g Cl2 ·
4.81
1 mol Cl2
2 mol FeCl3
162.2 g
·
·
= 15.2 g FeCl3
70.91 g
3 mol Cl2
1 mol FeCl3
(a) titanium(IV) chloride, water, titanium(IV) oxide, hydrogen chloride
(b) 14.0 mL TiCl4 ·
1.73 g 1 mol TiCl4
2 mol H2 O
18.02 g
·
·
·
= 4.60 g H2O
1 mL
189.7 g
1 mol TiCl4 1 mol H2 O
(c) 14.0 mL TiCl4 ·
1.73 g 1 mol TiCl4 1 mol TiO2
79.87 g
·
·
·
= 10.2 g TiO2
1 mL
189.7 g
1 mol TiCl4 1 mol TiO2
14.0 mL TiCl4 ·
4.82
(a) According to the text (p. 164), O2 is the limiting reactant.
750. g O2 ·
1 mol O2
6 mol H2 O
18.02 g
·
·
= 507 g H2O
32.00 g
1 mol H2 O
5 mol O2
(b) 750. g NH3 ·
4.83
1.73 g 1 mol TiCl4
36.46 g
4 mol HCl
·
·
·
= 18.6 g HCl
1 mL
189.7 g
1 mol TiCl4 1 mol HCl
1 mol NH3
5 mol O2
32.00 g
·
·
= 1760 g O2
17.03 g
4 mol NH3 1 mol O2
15.0 g NaNO3 ·
1 mol NaNO3
1 mol NaN3
65.01 g
·
·
= 11.5 g NaN3
84.99 g
1 mol NaNO3 1 mol NaN3
15.0 g NaNH2 ·
1 mol NaNH2
1 mol NaN3
65.01 g
·
·
= 8.33 g NaN3
39.01 g
3 mol NaNH2 1 mol NaN3
The maximum mass that can be produced is 8.33 g NaN3.
80
Chapter 4
4.84
Stoichiometry: Quantitative Information about Chemical Reactions
(a) Sodium iodate, sodium hydrogen sulfite (or sodium bisulfite)
103 g 1 mol I2
2 mol NaIO3
197.9 g
·
·
·
= 1560 g NaIO3
1 kg
253.8 g
1 mol NaIO3
1 mol I2
(b) 1.00 kg I2 ·
103 g 1 mol I2 5 mol NaHSO3
104.1 g
·
·
·
= 2050 g NaHSO3
1 kg
253.8 g
1 mol I2
1 mol NaHSO3
1.00 kg I2 ·
(c) 15.0 g NaIO3 ·
0.125 L ·
1 mol NaIO3
1 mol I2
253.8 g
·
·
= 9.62 g I2
197.9 g
2 mol NaIO3 1 mol I2
0.853 mol NaHSO3
1 mol I2
253.8 g
·
·
= 5.41 g I2
5 mol NaHSO3 1 mol I2
1L
The maximum mass that can be produced is 5.41 g I 2.
4.85
0.2070 g BaSO4 ·
1 mol BaSO4
1 mol saccharin
183.19 g
1 mol S
·
·
·
= 0.1625 g
1 mol S
1 mol saccharin
233.39 g
1 mol BaSO4
saccharin
0.1625 g
· 100% = 75.92% saccharin
0.2140 g
4.86
0.422 g B2O3 ·
1 mol B2 O3
10.81 g
2 mol B
·
·
= 0.131 g B
69.62 g
1 mol B2 O3 1 mol B
0.148 g BxHy – 0.131 g B = 0.017 g H
0.131 g B ·
1 mol B
= 0.0121 mol B
10.81 g
0.017 mol H
1.4 mol H
7 mol H
=
=
0.0121 mol B
1 mol B
5 mol B
4.87
1 mol SiO2
1 mol Si
·
= 0.1937 mol Si
60.084 g
1 mol SiO2
6.980 g H2O ·
1 mol H2 O
2 mol H
·
= 0.7749 mol H
18.015 g
1 mol H2 O
269 mg CO2 ·
The empirical formula is SiH4
1g
1 mol CO2
1 mol C
·
·
= 0.00611 mol C
3
10 mg
44.01 g
1 mol CO2
0.00611 mol C ·
111 mg H2O ·
1 mol H
= 0.017 mol H
1.01 g
The empirical formula is B5H7
11.64 g SiO2 ·
0.7749 mol H
4 mol H
=
0.1937 mol Si 1 mol Si
4.88
0.017 g H ·
12.01 g
= 0.0734 g C
1 mol C
1g
1 mol H2 O
2 mol H
·
·
= 0.0123 mol H
103 mg
18.0 g
1 mol H2 O
81
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
0.012 mol H ·
1.01 g
= 0.012 g H
1 mol H
mass of O = sample mass – mass of C – mass of H
= (95.6 mg ·
1g
) – 0.0734 g – 0.0123 g
103 mg
= 0.010 g O
0.010 g O ·
1 mol O
= 0.00063 mol O
16.0 g
0.00611 mol C 10 mol C
=
0.00063 mol O
1 mol O
0.0123 mol H
20 mol H
=
0.00063 mol O
1 mol O
The empirical formula is C10H20O. This problem is very sensitive to the proper use of significant figures.
4.89
0.257 g CO2 ·
1 mol CO2
1 mol C
·
= 0.00584 mol C
44.01 g
1 mol CO2
0.00584 mol C ·
12.01 g
= 0.0701 g C
1 mol C
0.0350 g H2O ·
1 mol H2 O
2 mol H
·
= 0.00388 mol H
18.02 g
1 mol H2 O
0.00388 mol H ·
1.008 g
= 0.00391 g H
1 mol H
mass of O = sample mass – mass of C – mass of H
= 0.105 g – 0.0701 – 0.00391
= 0.031 g O
0.031 g O ·
1 mol O
= 0.0019 mol O
16.0 g
0.00584 mol C
3 mol C
=
0.0019 mol O
1 mol O
The empirical formula is C3H2O
82
0.00388 mol H
2 mol H
=
0.0019 mol O
1 mol O
Chapter 4
4.90
Stoichiometry: Quantitative Information about Chemical Reactions
(a) FeCl2(aq) + Na2S(aq)  FeS(s) + 2 NaCl(aq)
(b) 40. g Na2S ·
40. g FeCl2 ·
1mol FeS
87.9 g
1 mol Na2 S
·
·
= 45 g FeS
1 mol Na 2 S 1 mol FeS
78.0 g
1 mol FeCl2
1mol FeS
87.9 g
·
·
= 28 g FeS
1 mol FeCl2 1 mol FeS
127 g
FeCl2 is the limiting reactant
(c) 28 g FeS
(d) 40. g FeCl2 ·
1 mol FeCl2
1mol Na 2 S
78.0 g
·
·
= 25 g Na2S required
1 mol FeCl2 1 mol Na2 S
127 g
40. g Na2S available – 25 g Na2S required = 15 g Na2S remains
(e) 40. g Na2S ·
1 mol Na 2S 1mol FeCl2
127 g
·
·
= 65 g FeCl2
1 mol Na2 S 1 mol FeCl2
78.0 g
Answers taken from the Simulation on Screen 4.8:
(a) FeCl2(aq) + Na2S(aq)  FeS(s) + 2 NaCl(aq)
(b) FeCl2
(c) 27.7 g
(d) 15.4 g Na2S
(e) 65 g
4.91
3.00 kg Cu2S ·
4.92
0.376 g CO2 ·
103 g 1 mol Cu2 S
1 mol H2SO4
1 mol S
98.08 g
·
·
·
·
= 1850 g H2SO4
1 kg
1 mol S
159.2 g
1 mol Cu2 S
1 mol H2SO4
1 mol CO2 1 mol MCO3
·
= 0.00854 mol MCO3
44.01 g
1 mol CO2
1.056 g
= 124 g/mol
0.00854 mol
124 g/mol (MCO3) – 60 g/mol (CO3) = 64 g/mol
4.93
The metal is (b) copper, Cu
M(s) + O2 (g)  MO2(s)
(0.452 – 0.356) g O2 ·
1 mol O2
1 mol M
·
= 0.0030 mol M
32.00 g
1 mol O2
0.356 g
= 120 g/mol (119 g/mol with three significant figures)
0.0030 mol
The metal is probably Sn (118.67 g/mol)
83
Chapter 4
4.94
Stoichiometry: Quantitative Information about Chemical Reactions
1.598 g TiO2 ·
1 mol TiO2
1 mol Ti
·
= 0.02001 mol Ti
79.866 g
1 mol TiO2
47.867 g
= 0.9577 g Ti
1 mol Ti
0.02001 mol Ti ·
1.438 g TixOy – 0.9577 g Ti = 0.480 g O
0.480 g O ·
1 mol O
= 0.0300 mol O
16.00 g
0.0300 mol O
3 mol O
=
0.02001 mol Ti
2 mol Ti
4.95
234 kg KClO4 ·
The empirical formula is Ti2O3
103 g 1 mol KClO4
1 mol Cl2
4 mol KClO3
3 mol KClO
70.91 g
·
·
·
·
·
1 kg
138.5 g
3 mol KClO4 1 mol KClO3 1 mol KClO 1 mol Cl2
= 4.79  105 g Cl2
4.96
(a) 125 kg Zn ·
103 g 1 mol Zn 1 mol ZnS2 O4 1 mol Na 2S2O4
174.1 g
·
·
·
·
1 kg
1 mol Zn
65.39 g
1 mol ZnS2 O4
1 mol Na2 S2 O4
= 3.33  105 g Na2S2O4
500. kg SO2 ·
103 g 1 mol SO2 1 mol ZnS2 O4 1 mol Na 2S2O4
174.1 g
·
·
·
·
1 kg
64.06 g
1 mol ZnS2 O4
1 mol Na2 S2 O4
2 mol SO2
= 1.36  106 g Na2S2O4
The maximum mass produced is 3.33  105 g Na2S2O4 (333 kg).
(b) 333 kg Na2S2O4 ·
4.97
125 kg limestone ·
100.0 kg commercial product
= 370. kg commercial product
90.1 kg Na2 S2 O4
103 g
95.0 g CaCO3
1 mol CaCO3
56.08 g
1 mol CaO
·
·
·
·
1 kg
100.0 g limestone
100.1 g
1 mol CaCO3 1 mol CaO
= 6.65 104 g CaO (66.5 kg)
4.98
8.63 g Ag ·
1 mol Ag 1 mol Ag2 MoS4
439.9 g
·
·
= 17.6 g Ag2MoS4
107.9 g
2 mol Ag
1 mol Ag2 MoS4
3.36 g Mo ·
1 mol Mo 1 mol Ag2 MoS4
439.9 g
·
·
= 15.4 g Ag2MoS4
1 mol Mo
95.94 g
1 mol Ag2 MoS4
4.81 g S ·
1 mol S 1 mol Ag2 MoS4
439.9 g
·
·
= 16.5 g Ag2MoS4
4 mol S
32.07 g
1 mol Ag2 MoS4
The maximum mass that can be obtained is 15.4 g Ag2MoS4
84
Chapter 4
4.99
Stoichiometry: Quantitative Information about Chemical Reactions
C4H8(g) + 6 O2(g)  4 CO2(g) + 4 H2O(g)
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
The problem can be solved using either the mass of CO2 or the mass of H2O. Set up two equations with two
unknowns.
x g C4H8 + y g C4H10 = 2.86 g
8.80 g CO2 ·
 x g C4 H8
1 mol CO2
4 mol CO2   y g C4 H10
8 mol CO2 
=


+

44.01 g
2 mol C4 H10 
 56.11 g/mol 1 mol C4 H8   58.12 g/mol
Substitute (2.86 g – x g C4H8) for y and solve.
y = 2.86 g – 1.27 g = 1.59 g C4H10
x = 1.27 g C4H8
1.27 g C4 H8
· 100% = 44.4% C4H8
2.86 g mixture
4.100
250 layers ·
100.0% – 44.4% = 55.6% C4H10
100 cm
0.60 nm
1m
· 9
·
= 1.5  105 cm thick
1m
1 layer
10 nm
2
 100 cm 
3
volume = (1.5  105 cm)(3.00 m)(3.00 m) 
 = 1.4 cm
 1m 
mass = 1.4 cm3 ·
4.101
1.0 g
1 cm3
100.0 g impure ore ·
75.4 g impure Cu ·
= 1.4 g (CH3)2SiCl2
89.0 g CuS/Cu2S mixture
= 89.0 g CuS/Cu2S mixture
100.0 g impure ore
89.5 g Cu
= 67.5 g Cu
100.0 g impure Cu
89.0 g mixture = x g CuS + y g Cu2S
 63.55 g Cu 
 127.09 g Cu 
67.5 g Cu = (x g CuS) 
 + (y g Cu2S) 

95.61
g
CuS


 159.16 g Cu 2S 
Substitute (89.0 – x) for y and solve.
y = 62.2 g Cu2S = 62.2% Cu2S
x = 26.8 g CuS = 26.8% CuS
4.102
In 0.015 M HCl, [H3O+] = 0.015 M
In pH 1.2 solution, [H3O+] = 10–pH = 10–1.2 = 0.06 M
The pH 1.2 solution has a higher hydronium ion concentration.
4.103
(a) MgCO3(s) + 2 H3O+(aq)  CO2(g) + Mg2+(aq) + 3 H2O()
(b) gas-forming reaction
(c) [H3O+] = 10–pH = 10–1.56 = 0.028 M
125 mL ·
0.028 mol H3 O+ 1 mol MgCO3
84.31 g
1L
·
·
·
= 0.15 g MgCO3
3
+
1L
10 mL
1 mol MgCO3
2 mol H3 O
85
Chapter 4
4.104
Stoichiometry: Quantitative Information about Chemical Reactions
H3C6H5O7(aq) + 3 NaHCO3(aq)  3 H2O() + 3 CO2(g) + Na3C6H5O7(aq)
0.100 g H3C6H5O7 ·
4.105
0.125 L ·
1 mol H3 C6 H5 O7
3 mol NaHCO3
84.01 g
·
·
= 0.131 g NaHCO3
192.1 g
1 mol H3 C6 H5 O7 1 mol NaHCO3
0.15 mol CH3CO2 H
1 mol NaHCO3
84.01 g
·
·
= 1.58 g NaHCO3
1L
1 mol CH3CO2 H 1 mol NaHCO3
15.0 g NaHCO3 is available, so CH3CO2H is the limiting reactant
0.125 L ·
0.15 mol CH3CO2 H 1 mol NaCH3CO2
82.03 g
·
·
= 1.54 g NaCH3CO2
1L
1 mol CH3CO2 H
1 mol NaCH3CO2
4.106
0.03351 L ·
0.0102 mol NaOH 1 mol H3 C6 H5 O7
192.13 g
·
·
= 0.219 g H3C6H5O7
3 mol NaOH
1L
1 mol H3 C6 H5 O7
4.107
0.04021 L ·
0.246 mol I2 2 mol Na2 S2 O3
158.11 g
·
·
= 3.13 g Na2S2O3
1L
1 mol I2
1 mol Na2 S2 O3
3.13 g
· 100% = 96.8%
3.232 g
4.108
0.02958 L ·
0.550 mol NaOH 1 mol H2 C2 O4
90.035 g
·
·
= 0.732 g H2C2O4
2 mol NaOH
1L
1 mol H2 C2 O4
0.732 g
· 100% = 16.1%
4.554 g
4.109
(a) pH = –log[H3O+] = –log(0.105) = 0.979
(b) [H3O+] = 10–pH = 10–2.56 = 0.0028 M
The solution is acidic
(c) [H3O+] = 10–pH = 10–9.67 = 2.1  10–10 M
The solution is basic
(d) cd = cc ·
10.0 mL
Vc
= 2.56 M ·
= 0.102 M
250. mL
Vd
pH = –log[H3O+] = –log(0.102) = 0.990
4.110
Net ionic equation: H3O+(aq) + HCO3–(aq)  2 H2O() + CO2(g)
[H3O+] = 10–pH = 10–2.56 = 0.0028 M H3O+
4.111
0.125 L ·
0.0028 mol H3O+ 1 mol NaHCO3
84.01 g
·
·
= 0.029 g NaHCO3
+
1L
1 mol H3 O
1 mol NaHCO3
0.500 L ·
2.50 mol HCl
3.75 mol HCl
+ 0.250 L ·
= 2.19 mol HCl
1L
1L
2.19 mol HCl
= 2.92 M HCl
0.750 L
pH = –log[H3O+] = –log(2.92) = –0.465
86
Chapter 4
4.112
Stoichiometry: Quantitative Information about Chemical Reactions
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O()
[H3O+] = 10–1.92 = 0.012 M H3O+ = 0.012 M HCl
0.250 L ·
0.250 L ·
0.012 mol
= 0.0030 mol HCl
1L
0.0105 mol NaOH
1 mol HCl
·
= 0.00263 mol HCl reacted
1
mol NaOH
1L
0.0030 mol – 0.00263 mol = 0.0004 mol HCl remains
 0.0004 mol 
pH = –log 
 = 3.1
 0.500 L 
4.113
0.250 L ·
0.125 mol HCl 1 mol CaCO3
100.1 g
·
·
= 1.56 g CaCO3 required
2 mol HCl
1L
1 mol CaCO3
2.56 g available – 1.56 g required = 1.00 g CaCO3 remains
0.250 L ·
4.114
0.125 mol HCl 1 mol CaCl2
111.0 g
·
·
= 1.73 g CaCl2
2 mol HCl
1L
1 mol CaCl2
(a) (NH4)2PtCl4(aq) + 2 NH3(aq)  Pt(NH3)2Cl2(aq) + 2 NH4Cl(aq)
(b) 12.50 g Pt(NH3)2Cl2 ·
1 mol Pt(NH3 )2 Cl2 1 mol (NH4 )2 PtCl4
372.97 g
·
·
300.05 g
1 mol Pt(NH3 )2 Cl2 1 mol (NH4 )2 PtCl4
= 15.54 g (NH4)2PtCl4
12.50 g Pt(NH3)2Cl2 ·
(c) 0.0370 L ·
1.50 mL ·
1 mol Pt(NH3 )2 Cl2
2 mol NH3
1L
·
·
= 0.667 L solution
300.05 g
1 mol Pt(NH3 )2 Cl2
0.125 mol NH3
0.475 mol HCl 1 mol C5 H5 N
·
= 0.0176 mol C5H5N unused by titration
1 mol HCl
1L
0.979 g 1 mol C5 H5 N
·
= 0.0186 mol C5H5N originally added
1 mL
79.10 g
0.0186 mol – 0.0176 mol = 0.0010 mol C5H5N reacted with cisplatin
0.150 g Pt(NH3)2Cl2 ·
1 mol Pt(NH3 )2 Cl2
= 5.00  10–4 mol Pt(NH3)2Cl2
300.0 g
0.0010 mol C5 H5 N
2 mol C5 H5 N
=
–4
5.00  10 mol Pt(NH3 )2 Cl2
1 mol Pt(NH3 )2 Cl2
The compound formula is Pt(NH3)2Cl2(C5H5N)2
4.115
1.0 g ·
Vd =
4.116
1 mol methylene blue
1
·
= 0.063 M methylene blue
0.0500 L
320. g
0.063 M
cc
· Vc =
· 0.0500 L = 7.6  104 L
4.1  10–8 M
cd
0.125 L ·
2.55 mol HCl 1 mol metal oxide
·
= 0.159 mol metal oxide
2 mol HCl
1L
Set up two equations with two unknowns.
87
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
0.159 mol metal oxide = (x g CaO ·
1 mol CaO
1 mol MgO
) + (y g MgO ·
)
40.30 g
56.08 g
7.695 g metal oxide = x g CaO + y g MgO
Substitute (7.695 g – y g MgO) for x and solve.
y = 3.17 g MgO
x = 4.52 g CaO
4.52 g CaO
· 100% = 58.7% CaO
7.695 g sample
4.117
3.17 g MgO
· 100% = 41.2% MgO
7.695 g sample
(a) Au, gold, has been oxidized and is the reducing agent
O2, oxygen, has been reduced and is the oxidizing agent
(b) 103 kg ore ·
4.118
103 g 0.019 g Au 1 mol Au 8 mol NaCN
1L
·
·
·
·
= 26 L solution
1 kg
4 mol Au
0.075 mol NaCN
100. g ore
197 g
FeCl3(aq) + 3 NaOH(aq)  Fe(OH)3(s) + 3 NaCl(aq)
(a) 0.0250 L ·
0.234 mol FeCl3 1 mol Fe(OH)3
106.9 g
·
·
= 0.625 g Fe(OH)3
1 mol Fe(OH)3
1L
1 mol FeCl3
0.0425 L ·
0.453 mol NaOH 1 mol Fe(OH)3
106.9 g
·
·
= 0.686 g Fe(OH)3
3 mol NaOH
1 mol Fe(OH)3
1L
NaOH is the excess reactant and 0.625 g Fe(OH)3 precipitates
(b) 0.0250 L ·
0.0425 L ·
0.234 mol FeCl3 3 mol NaOH
·
= 0.0176 mol NaOH required
1L
1 mol FeCl3
0.453 mol NaOH
– 0.0176 mol NaOH = 0.0017 mol NaOH remains
1L
0.0017 mol
= 0.025 M NaOH
0.0675 L
4.119
% atom economy = M(CH3CH2CH2CH2Br)/[M(CH3CH2CH2CH2OH) + M(NaBr) + M(H2SO4)] · 100%
137.02 g/mol/(74.12 g/mol + 102.89 g/mol + 98.08 g/mol) · 100% = 49.81%
4.120
(a) “chlorohydrin route”: % atom economy = M(C2H4O)/[M(C2H4) + M(Cl2) + M(Ca(OH)2)] · 100%
44.05 g/mol/(28.05 g/mol + 70.90 g/mol + 74.09 g/mol) · 100% = 25.46%
88
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
For modern catalytic reaction, all atoms in reactants are in product. % atom economy is 100%. The
modern catalytic reaction is more efficient.
(b) Theoretical yield = 867 g C2H4 ·
1 mol C 2 H 4
·
1 mol C 2 H 4 O
28.05 g C 2 H 4
1 mol C 2 H4
·
44.04 g C 2 H 4 O
= 1.36 x 103 g C2H4O
1 mol C 2 H 4 O
762 g/ 1.36 x 103 g · 100% = 56.0 % yield
4.121
4.122
cd = c c ·
25.0 mL
Vc
= 0.110 M ·
= 0.0275 M
100.0 mL
Vd
0.0275 M ·
10.0 mL
= 0.00110 M Na2CO3
250. mL
0.0500 L ·
0.100 mol HCl
= 0.00500 mol HCl (total)
1L
0.0111 L ·
0.121 mol NaOH
1 mol HCl
·
= 0.00134 mol HCl (excess)
1L
1 mol NaOH
0.00500 mol HCl (total) – 0.00134 mol HCl (excess) = 0.00366 mol HCl (reacted with NH 3)
0.00366 mol HCl ·
1 mol NH3 1 mol (NH4 )2 SO4
132.1 g
= 0.242 g (NH4)2SO4
·
·
1 mol HCl
2 mol NH3
1 mol (NH4 )2 SO4
0.242 g (NH4 )2SO4
· 100% = 50.8%
0.475 g sample
4.123
(a) Reaction 1: Cu2+ is reduced and is the oxidizing agent; I– is oxidized and is the reducing agent
Reaction 2: I3– is reduced and is the oxidizing agent; S2O32– is oxidized and is the reducing agent
(b) 0.02632 L ·
0.101 mol Na 2S2 O3
1 mol I3–
·
= 0.00133 mol I3–
1L
2 mol Na2 S2 O3
0.00133 mol I3– ·
63.55 g
2 mol Cu 2+
·
= 0.169 g Cu
–
1 mol Cu 2+
1 mol I3
0.169 g
· 100% = 67.3%
0.251 g
4.124
0.03450 L ·
0.108 mol KMnO4
5 mol C2 O42–
·
= 0.00932 mol C2O42–
1L
2 mol KMnO4
Use mol C2O42– to determine which formula is correct:
0.00932 mol C2O42– ·
307.0 g
1 mol K[Fe(C2 O4 )2 (H2O)2 ]
·
= 1.43 g compound
2–
1 mol K[Fe(C2 O4 )2 (H2 O)2 ]
2 mol C2 O4
0.00932 mol C2O42– ·
437.2 g
1 mol K3[Fe(C2O4 )3 ]
·
= 1.36 g compound
2–
1 mol K3 [Fe(C2 O4 )3 ]
3 mol C2 O4
The correct formula is K3[Fe(C2O4)3]
89
Chapter 4
4.125
Stoichiometry: Quantitative Information about Chemical Reactions
1.500 mol HCl 1 mol NH4
1 mol NH3
·
·
= 0.03639 mol NH3
1L
1 mol NH4
1 mol HCl
0.02426 L ·
0.03639 mol NH3 ·
17.030 g
= 0.6197 g NH3
1 mol NH3
1.580 g compound – 0.6197 g NH3 = 0.960 g CrCl3
0.960 g CrCl3 ·
4.126
1 mol CrCl3
= 0.00606 mol CrCl3
158.4 g
0.301 g BaSO4 ·
x=
0.03545 mol NH3
=6
0.00606 mol CrCl3
103 mg
1 mol thioridazine
370.6 g
1 mol BaSO4
1 mol S
·
·
·
·
1g
2 mol S
1 mol thioridazine
233.4 g
1 mol BaSO4
= 239 mg thioridazine
239 mg
= 19.9 mg thioridazine per tablet
12 tablets
4.127
0.1840 g AgCl ·
1 mol AgCl
1 mol Cl
1 mol 2,4-D
221.04 g
·
·
·
= 0.1419 g 2,4-D
2 mol Cl
143.32 g
1 mol AgCl
1 mol 2,4-D
0.1419 g
· 100% = 11.48% 2,4-D
1.236 g
4.128
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O()
(a) Average volume (NaOH sample) =
0.02055 L ·
cc = c d ·
(b)
4.129
(20.15 + 21.30 + 20.40 + 20.35) mL
= 20.55 mL
4
0.1760 mol NaOH 1mol H2SO4
1
·
·
= 0.1808 M H2SO4 (diluted sample)
2 mol NaOH 0.01000 L
1L
500. mL
Vd
= 0.1808 M ·
= 18.1 M H2SO4 (concentrated sample)
5.00 mL
Vc
18.1 mol H2SO4 98.08 g H2SO4 1.00 L solution
1 kg
·
·
·
· 100% = 96.5%
1.00 L solution
1 mol H2SO4
1.84 kg solution 103 g
80 g H2O ·
74.5 g CaCl2
= 59.6 g CaCl2 dissolved in solution
100 g H2 O
74.9 g CaCl2·6H2O ·
1 mol CaCl2
111.0 g
1 mol CaCl2 ·6H2 O
·
·
= 37.9 g CaCl2 precipitated
1 mol CaCl2 ·6H2 O 1 mol CaCl2
219.1 g
(59.6 + 37.9) g CaCl2 ·
1 mol CaCl2
= 0.878 mol CaCl2
111.0 g
150 g solid – 97.5 g CaCl2 = 53 g H2O
53 g ·
1 mol H2 O
= 2.9 mol H2O
18.02 g
2.9 mol H2 O
= 3.3 mol H2O/mol CaCl2
0.878 mol CaCl2
90
Chapter 4
4.130
Stoichiometry: Quantitative Information about Chemical Reactions
5 Fe2+(aq) + MnO4–(aq) + 8 H3O+(aq)  Mn2+(aq) + 5 Fe3+(aq) + 12 H2O()
0.04240 mol MnO4–
1 mol Fe
5 mol Fe2+
·
·
= 0.007950 mol Fe in original mixture
–
1L
1 mol Fe2+
1 mol MnO4
0.03750 L ·
Set up two equations with two unknowns:
 1 mol Fe 
 1 mol Fe2 O3   2 mol Fe 
0.007950 mol Fe = (x g Fe) 
 + (y g Fe2O3) 


55.845
g
Fe


 159.69 g Fe2 O3   1 mol Fe2 O3 
0.551 g mixture = x g Fe + y g Fe2O3
Substitute (0.551 – y) for x and solve.
y = 0.356 g Fe2O3
mass percent =
0.356 g Fe2 O3
· 100% = 64.6% Fe2O3
0.551 g mixture
x = 0.195 g Fe
mass percent =
0.195 g Fe
· 100% = 35.4% Fe
0.551 g mixture
4.131
Absorbance
1
0.8
0.6
0.4
y = 205600x + 0.024
0.2
0
0
2E-06
4E-06
6E-06
g P/L
(a) slope = 2.06  105; intercept = 0.0240
(b) 0.518 = 2.06  105(diluted urine solution concentration) + 0.0240
diluted urine solution concentration = 2.40  10–6 g/L
cc = c d ·
50.00 mL
Vd
= 2.40  10–6 g/L ·
= 1.20  10–4 g/L
1.00 mL
Vc
(c) 1.20  10–4 g/L · 1.122 L/day = 1.35  10–4 g P/day
(1.35 x 10-4 g P/day)(94.97 g PO43-/30.97 g P) = 4.13 x 10-4 g PO43-/day
4.132
Mohr titration: Ag+(aq) + Cl–(aq)  AgCl(s)
0.04100 L ·
1 mol Cl–
0.0750 mol Ag+
1
·
·
= 0.0615 M Cl– (as KCl) in mixture
1L
1 mol Ag+
0.05000 L
0.03812 L ·
1 mol Cl–
0.0750 mol Ag+
1
·
·
= 0.114 M Cl– (total) in mixture
+
1L
1 mol Ag
0.02500 L
0.114 M Cl– (total) – 0.0615 M Cl– (as KCl) = 0.053 M KClO4
0.250 L solution ·
74.55 g
0.0615 mol KCl
·
= 1.15 g KCl
1 mol KCl
1L
91
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
0.250 L solution ·
0.053 mol KClO4
138.5 g
·
= 1.8 g KClO4
1 mol KClO4
1L
1.15 g KCl
· 100% = 28.8 % KCl
4.000 g mixture
4.133
1.8 g KClO4
· 100% = 45% KClO4
4.000 g mixture
The total mass of the beakers and solutions after reaction is equal to the total mass before the reaction
(167.170 g) because no gases were produced in the reaction.
4.134
(a) 10.8 g product – 2.0 g Fe = 8.8 g Br2
(b) 2.0 g Fe ·
1 mol Fe
= 0.036 mol Fe
55.8 g
8.8 g Br2 ·
1 mol Br2
2 mol Br
·
= 0.11 mol Br
160. g
1 mol Br2
0.11 mol Br
3 mol Br
=
0.036 mol Fe 1 mol Fe
(c) FeBr3
(d) 2 Fe(s) + 3 Br2()  2 FeBr3(s)
(e) Iron(III) bromide
(f) (i) When 1.00 g of Fe is added to the Br 2, Fe is the limiting reagent.
4.135
A solution containing 0.100 mol of HCl requires 0.0500 mol of Zn or 3.27 g for complete reaction.
0.100 mol HCl ·
1 mol Zn
65.4 g
·
= 3.27 g Zn
2 mol HCl 1 mol Zn
Thus, in flask 2 the reaction just uses all of the Zn and produces 0.050 mol H 2 gas.
0.100 mol HCl ·
1 mol H 2
= 0.050 mol H2
2 mol HCl
In flask 1, containing 7.00 g of Zn, some Zn remains after the HCl has been consumed; 0.050 mol H2 gas is
produced. In flask 3, there is insufficient Zn, so less hydrogen is produced.
4.136
92
Some aluminum remains in the beaker. Bromine was the limiting reactant.
Chapter 4
4.137
Stoichiometry: Quantitative Information about Chemical Reactions
If both students base their calculations on the amount of HCl solution pipetted into the flask (20.0 mL), then
the second student’s result will be (e) the same as the first student’s. However, if the HCl concentration is
calculated using the diluted solution volume, student 1 will use a volume of 40.0 mL and student 2 will use
a volume of 80.0 mL in the calculation. The second student’s result will be (c) two times less than the first
student’s.
4.138
103 mg
0.033 mol C2 H5OH
1L
46.1 g
·
·
·
= 150 mg/dL
1g
10 dL
1L
1 mol C2 H5OH
The person is intoxicated and will be arrested.
4.139
(a) % atom economy = M(C4H2O3)/[M(C6H6) + 4.5M(O2)] · 100%
(98.06 g/mol)/(78.11 g/mol + 4.5 · 32.00 g/mol) · 100% = 44.15%
(b) Theoretical yield =
1 mol C 6 H 6
1.00 x 103 g C6H6 ·
·
78.11 g C 6 H 6
1 mol C 4 H 2 O 3
·
1 mol C 6 H 6
98.06 g C 4 H 2 O 3
= 1.26 x 103 g C4H2O3
1 mol C 4 H 2 O 3
Percent yield = 972 g/1.26 x 103 g · 100% = 77.1 %
1 mol C 6 H 6
1.0 x 103 g C6H6 ·
78.11 g C 6 H 6
·
2 mol CO 2
·
44.00 g CO2
1 mol C 6 H 6
= 1.13 x 103 g CO2 = 1.13 kg CO2
1 mol CO2
SOLUTIONS TO APPLYING CHEMICAL PRINCIPLES: ANTACIDS
1.
NaHCO3, KHCO3, and CaCO3
2.
(a) CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O() + CO2(g)
(b) 0.500 g CaCO3(1 mol CaCO3/100.09 g)(2 mol HCl/1 mol CaCO3)(1 L/0.500 mol HCl) = 0.0200 L =
20.0 mL
3.
(a) Mg(OH)2(s) + 2 HCl(aq)  MgCl2(aq) + 2 H2O()
(b) 0.550 g CaCO3(1 mol CaCO3/100.09 g)(2 mol HCl/1 mol CaCO3)(1 L/0.500 mol HCl) = 0.02198 L =
21.98 mL
29.52 mL – 21.98 mL = 7.54 mL HCl required to titrate the Mg(OH)2
0.00754 L(0.500 mol HCl/1 L)(1 mol Mg(OH) 2/2 mol HCl)(58.32 g Mg(OH)2/ 1 mol) = 0.110 g Mg(OH)2
= 1.10 x 102 mg Mg(OH)2
4.
0.200 g Mg(OH)2(1 mol Mg(OH)2/58.32 g Mg(OH)2)(2 mol HCl/mol Mg(OH)2)(1 L/0.500 mol HCl) =
0.01372 L or 13.72 mL
93
Chapter 4
Stoichiometry: Quantitative Information about Chemical Reactions
0.200 g Al(OH)3(1 mol Al(OH)3/78.00 g Al(OH)3)(3 mol HCl/mol Al(OH)3)(1 L/0.500 mol HCl) =
0.01538 L or 15.38 mL
13.72 mL + 15.38 mL = 29.10 mL of 0.500 M HCl solution
5.
A tablet of Rolaids. 29.52 mL of 0.500 HCl is required for a tablet of Rolaids compared to 29.10 mL for a
teaspoon of Maalox and 20.00 mL for a tablet of Tums.
94