CHAPTER 4
CHEMICAL REACTIONS
PRACTICE EXAMPLES
1A
(a)
Unbalanced reaction:
Balance Ca & PO43-:
Balance H atoms:
Self Check:
H3PO4(aq) + CaO(s)
2 H3PO4(aq) + 3 CaO(s)
2 H3PO4(aq) + 3 CaO(s)
6 H + 2 P + 11 O + 3 Ca




Ca3(PO4)2(aq) + H2O(l)
Ca3(PO4)2(aq) + H2O(l)
Ca3(PO4)2(aq) + 3 H2O(l)
6 H + 2 P + 11 O + 3 Ca
(b)
Unbalanced reaction:
Balance C& H:
Balance O atoms:
Self Check:
C3H8(g) + O2(g)
C3H8(g) + O2(g)
C3H8(g) + 5 O2(g)
3 C + 8 H + 10 O




CO2(g) + H2O(g)
3 CO2(g) + 4 H2O(g)
3 CO2(g) + 4 H2O(g)
3 C + 8 H + 10 O
(a)
Unbalanced reaction:
NH3(g) + O2(g)
Balance N and H:
NH3(g) + O2(g)
Balance O atoms:
NH3(g) + 7/4 O2(g)
Multiply by 4 (whole #): 4 NH3(g) + 7 O2(g)
Self Check:
4 N + 12 H + 14 O





NO2(g) + H2O(g)
NO2(g) + 3/2 H2O(g)
NO2(g) + 3/2 H2O(g)
4 NO2(g) + 6 H2O(g)
4 N + 12 H + 14 O
(b)
Unbalanced reaction:
Balance H atoms:
Balance O atoms:
Balance N atoms:
Multiply by 4 (whole #)
Self Check:
NO2(g) + NH3(g)
NO2(g) + 2 NH3(g)
3/2 NO2(g) + 2 NH3(g)
3/2 NO2(g) + 2 NH3(g)
6 NO2(g) + 8 NH3(g)
14 N + 24 H + 12 O






N2(g) + H2O(g)
N2(g) + 3 H2O(g)
N2(g) + 3 H2O(g)
7/4 N2(g) + 3 H2O(g)
7 N2(g) + 12 H2O(g)
14 N + 24 H + 12 O
HgS(s) + CaO(s)
HgS(s) + 4 CaO(s)
HgS(s) + 4 CaO(s)
4 HgS(s) + 4 CaO(s)
4 HgS(s) + 4 CaO(s)
4 Hg + 4 S + 4 O + 4 Ca






1B
2A
Unbalanced reaction:
Balance O atoms:
Balance Ca atoms:
Balance S atoms:
Balance Hg atoms:
Self Check:
66
CaS(s) + CaSO4(s) + Hg(l)
CaS(s) + CaSO4(s) + Hg(l)
3 CaS(s) + CaSO4(s) + Hg(l)
3 CaS(s) + CaSO4(s) + Hg(l)
3 CaS(s) + CaSO4(s) + 4 Hg(l)
4 Hg + 4 S + 4 O + 4 Ca
Chapter 4: Chemical Reactions
2B
 CO2(g) + H2O(l) + SO2(g)
C7H6O2S(l) + O2 (g)
C7H6O2S(l) + O2(g)
 7 CO2(g) + H2O(l) + SO2(g)
C7H6O2S(l) + O2(g)
 7 CO2(g) + H2O(l) + SO2(g)
 7 CO2(g) + 3 H2O(l) + SO2(g)
C7H6O2S(l) + O2(g)
C7H6O2S(l) + 8.5 O2 (g)  7 CO2(g) + 3 H2O(l) + SO2(g)
2 C7H6O2S(l) + 17 O2 (g)  14 CO2(g) + 6 H2O(l) + 2 SO2(g)
14 C + 12 H + 2 S + 38 O 14 C + 12 H + 2 S + 38 O
Unbalanced reaction:
Balance C atoms:
Balance S atoms:
Balance H atoms:
Balance O atoms:
Multiply by 2 (whole #):
Self Check:
3A
The balanced chemical equation provides the factor needed to convert from moles KClO 3
3 mol O 2
to moles O2. Amount O 2 = 1.76 mol KClO 3 
= 2.64 mol O 2
2 mol KClO 3
3B
First, find the molar mass of Ag 2 O .
 2 mol Ag  107.87 g Ag  +16.00 g O = 231.74 g Ag 2 O / mol
amount Ag = 1.00 kg Ag 2 O 
4A
The balanced chemical equation provides the factor to convert from amount of Mg to
amount of Mg 3 N 2 . First we must determine the molar mass of Mg 3 N 2 .
molar mass =  3mol Mg  24.305g Mg  +  2 mol N  14.007 g N  = 100.93g Mg 3 N 2
mass Mg 3 N 2 = 3.82g Mg 
4B
1000 g
1 mol Ag 2 O
2 mol Ag


= 8.63 mol Ag
1.00 kg 231.74 g Ag 2 O 1 mol Ag 2 O
1mol Mg 1mol Mg 3 N 2 100.93g Mg 3 N 2


= 5.29 g Mg 3 N 2
24.31g Mg
3mol Mg
1mol Mg 3 N 2
The pivotal conversion is from H 2  g  to CH 3OH (l). For this we use the balanced equation,
which requires that we use the amounts in moles of both substances. The solution involves
converting to and from amounts, using molar masses.
2 mol H 2
2.016 g H 2
1000 g 1mol CH 3 OH
mass H 2  g  = 1.00 kg CH 3 OH(l) 



1 kg 32.04 g CH 3 OH 1mol CH 3 OH
1mol H 2
mass H 2  g  = 126 g H 2
5A
The equation for the cited reaction is: 2 NH 3  g  + 1.5 O 2  g  
 N 2 (g) + 3H 2 O  l 
The pivotal conversion is from one substance to another, in moles, with the balanced
chemical equation providing the conversion factor.
2 mol NH 3 17.0305g NH 3
1mol O 2
mass NH 3  g  = 1.00 g O 2  g  
= 0.710 g NH 3


32.00 g O 2 1.5 mol O 2
1mol H 2
67
Chapter 4: Chemical Reactions
5B
6A
6B
7A
25
O 2  g   8 CO 2  g  + 9 H 2 O  l 
2
1mol C8 H18
12.5 mol O 2 32.00 g O 2
mass O 2 = 1.00 g C8 H18 
= 3.50 g O 2  g 


114.23g C8 H18 1mol C8 H18 1mol O 2
The equation for the combustion reaction is: C8 H18  l  +
We must convert mass H 2  amount of H 2  amount of Al  mass of Al  mass of
alloy  volume of alloy. The calculation is performed as follows: each arrow in the
preceding sentence requires a conversion factor.
1mol H 2
2 mol Al 26.98g Al 100.0 g alloy 1cm3alloy
Valloy  1.000 g H 2 




2.016 g H 2 3mol H 2
1mol Al
93.7 g Al
2.85g alloy
Volume of alloy = 3.34 cm3 alloy
In the example, 0.207 g H 2 is collected from 1.97 g alloy; the alloy is 6.3% Cu by mass.
This information provides the conversion factors we need.
1.97 g alloy
6.3 g Cu
mass Cu = 1.31g H 2 

= 0.79 g Cu
0.207 g H 2 100.0 g alloy
Notice that we do not have to consider each step separately. We can simply use values
produced in the course of the calculation as conversion factors.
The cited reaction is 2 Al  s  + 6 HCl  aq   2 AlCl3  aq  + 3 H 2  g  . The HCl(aq) solution
has a density of 1.14 g/mL and contains 28.0% HCl. We need to convert between the
substances HCl and H 2 ; the important conversion factor comes from the balanced chemical
equation. The sequence of conversions is: volume of HCl(aq)  mass of HCl(aq)  mass
of pure HCl  amount of HCl  amount of H 2  mass of H 2 .
In the calculation below, each arrow in the sequence is replaced by a conversion factor.
1.14 g sol 28.0 g HCl 1mol HCl
3mol H 2 2.016 g H 2




mass H 2 = 0.05 mL HCl  aq  
1mL soln 100.0 g soln 36.46 g HCl 6 mol HCl 1mol H 2
mass H 2 = 4  104 g H 2  g  = 0.4 mg H 2  g 
7B
Density is necessary to determine the mass of the vinegar, and then the mass of acetic acid.
mass CO 2 (g) = 5.00 mL vinegar×
1mol CO2
44.01g CO 2
1.01g 0.040g acid 1mol CH3COOH
×
×
×
×
1mL 1g vinegar 60.05g CH3COOH 1mol CH3COOH 1mol CO2
= 0.15g CO2
8A
Determine the amount in moles of acetone and the volume in liters of the solution.
22.3g  CH 3 2 CO×
molarity of acetone =
8B
1mol  CH 3 2 CO
58.08g  CH 3 2 CO
1.25 L soln
= 0.307 M
The molar mass of acetic acid, HC2 H 3O 2 , is 60.05 g/mol. We begin with the quantity of
acetic acid in the numerator and that of the solution in the denominator, and transform to
the appropriate units for each.
68
Chapter 4: Chemical Reactions
molarity =
15.0 mL HC 2 H 3O 2 1000 mL 1.048g HC 2 H 3O 2 1mol HC 2 H 3O 2



= 0.524 M
500.0 mL soln
1L soln
1mL HC 2 H 3O 2
60.05g HC 2 H 3O 2
9A
The molar mass of NaNO 3 is 84.99 g/mol. We recall that “M” stands for “mol /L soln.”
10.8 mol NaNO3 84.99 g NaNO3
1L
mass NaNO3 = 125 mL soln 


= 115 g NaNO3
1000 mL
1L soln
1mol NaNO3
9B
We begin by determining the molar mass of Na 2SO 4  10H 2 O . The amount of solute needed
is computed from the concentration and volume of the solution.
mass Na 2SO 4  10H 2 O = 355 mL soln 

1L
1000 mL

322.21 g Na 2SO 4  10H 2 O
1 mol Na 2SO 4  10H 2 O
0.445 mol Na 2SO 4
1 L soln

1 mol Na 2SO 4  10H 2 O
1 mol Na 2SO 4
 50.9 g Na 2SO 4  10H 2 O
10A The amount of solute in the concentrated solution doesn’t change when the solution is
diluted. We take advantage of an alternative definition of molarity to answer the question:
millimoles of solute/milliliter of solution.
0.450 mmol K 2 CrO 4
amount K 2 CrO 4 = 15.00 mL 
= 6.75 mmol K 2 CrO 4
1 mL soln
6.75 mmol K 2 CrO 4
K 2 CrO 4 molarity, dilute solution =
= 0.0675 M
100.00 mL soln
10B We know the initial concentration (0.105 M) and volume (275 mL) of the solution, along
with its final volume (237 mL). The final concentration equals the initial concentration
times a ratio of the two volumes.
V
275mL
cf  ci  i  0105
. M
 0122
.
M
Vf
237 mL
11A The balanced equation is K 2 CrO 4  aq   2 AgNO3  aq   Ag 2 CrO 4  s   2 KNO3  aq  .
The molar mass of Ag 2 CrO 4 is 331.73 g mol . The conversions needed are mass
Ag 2 CrO 4  amount Ag 2 CrO 4 (moles)  amount K 2 CrO 4 (moles)  volume K 2 CrO 4  aq  .
VK 2CrO4 1.50 g Ag 2 CrO 4 
1mol Ag 2 CrO 4
331.73g Ag 2CrO 4
1000 mL solution

=18.1mL
1L solution

1mol K 2CrO 4
1L soln

1 mol Ag 2 CrO 4 0.250 mol K 2 CrO 4
11B Balanced reaction: 2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2 KNO3(aq)
moles of K2CrO4 = C  V = 0.0855 M  0.175 L sol = 0.01496 moles K2CrO4
2 mol AgNO3
= 0.0299 mol AgNO3
moles of AgNO3 = 0.01496 mol K2CrO4 
1 mol K 2 CrO 4
69
Chapter 4: Chemical Reactions
0.0299 mol AgNO3
n
= 0.1995 L or 2.00  102 mL (0.200 L) of AgNO3
=
mol
C 0.150
AgNO3
L
1mol Ag 2 CrO 4 331.73 g Ag 2 CrO 4
Mass of Ag2CrO4 formed = 0.01496 moles K2CrO4 

1 mol K 2 CrO 4
1mol Ag 2 CrO 4
Mass of Ag2CrO4 formed = 4.96 g Ag2CrO4
VAgNO3 =
12A Reaction: P4  s   6 Cl2  g   4 PCl3  l  . We must determine the mass of PCl 3 formed by
each reactant.
1 mol P4
4 mol PCl 3 137.33 g PCl 3


 953 g PCl 3
123.90g P4
1mol P4
1 mol PCl 3
1 mol Cl 2 4 mol PCl 3 137.33 g PCl 3
mass PCl 3  725 g Cl 2 


 936 g PCl 3
70.91g Cl 2 6 mol Cl 2
1 mol PCl 3
Thus, a maximum of 936g PCl 3 can be produced; there is not enough Cl 2 to produce any
more.
mass PCl 3  215 g P4 
12B Since data are supplied and the answer is requested in kilograms (thousands of grams),
we can use kilomoles (thousands of moles) to solve the problem. We calculate the amount
in kilomoles of POCl 3 that would be produced if each of the reactants were completely
converted to product. The smallest of these amounts is the one that is actually produced
(this is a limiting reactant question).
1kmol PCl3 10 kmol POCl3
amount POCl3 1.00 kg PCl3 

 0.0121kmol POCl3
137.33kg PCl3
6 kmol PCl3
1kmol Cl2
10 kmol POCl3
amount POCl3 1.00 kg Cl2 

 0.0235 kmol POCl3
70.905 kg Cl2
6 kmol Cl2
1kmol P4 O10
10 kmol POCl3
amount POCl3 1.00 kg P4 O10 

 0.0352 kmol POCl3
283.89 kg P4 O10 1kmol P4 O10
Thus, a maximum of 0.0121 kmol POCl3 can be produced.
We next determine the mass of the product.
153.33kg POCl3
mass POCl3  0.0121kmol POCl3 
1.86 kg POCl3
1 kmol POCl3
13A The 725 g Cl 2 limits the mass of product formed. The P4  s  therefore is the reactant in
excess. From the quantity of excess reactant we can find the amount of product formed:
953 g PCl 3  936 g PCl 3 = 17 g PCl 3 . We calculate how much P4 this is, both in the
traditional way and by using the initial  215 g P4  and final  953 g PCl3  values of the
previous calculation.
mass P4  17 g PCl 3 
1 mol PCl 3
1 mol P4
123.90 g P4


 38
. g P4
137.33g PCl 3 4 mol PCl 3
1 mol P4
70
Chapter 4: Chemical Reactions
13B Find the amount of H2O(l) formed by each reactant, to determine the limiting reactant.
1mol H 2
2 mol H 2 O
amount H 2 O  12.2 g H 2 

 6.05 mol H 2 O
2.016 g H 2 2 mol H 2
1mol O 2
2 mol H 2 O
amount H 2 O  154 g O 2 

 9.63 mol H 2 O
32.00 g O 2 1 mol O 2
Since H 2 is limiting, we must compute the mass of O 2 needed to react with all of the H 2
1mol O 2 32.00 g O 2
mass O 2 reacting = 6.05 mol H 2 O produced 

 96.8g O 2 reacting
2 mol H 2 O 1mol O 2
mass O 2 remaining = 154 g originally present  96.8g O 2 reacting =57 g O 2 remaining
14A
The theoretical yield is the calculated maximum mass of product expected if we were
to assume that the reaction has no losses (100% reaction).
1mol CH 2 O 30.03g CH 2 O
mass CH 2 O  g   1.00 mol CH 3OH 

 30.0 g CH 2 O
1mol CH 3OH 1mol CH 2 O
(b) The actual yield is what is obtained experimentally: 25.7 g CH2O (g).
(c) The percent yield is the ratio of actual yield to theoretical yield, multiplied by 100%:
25.7 g CH 2O produced
% yield =
 100 % = 85.6 % yield
30.0 g CH 2 O calculated
(a)
14B First determine the mass of product formed by each reactant.
1mol P4
4 mol PCl3 137.33g PCl3


mass PCl3 = 25.0 g P4 
= 111g PCl3
123.90 g P4
1mol P4
1mol PCl3
1mol Cl2 4 mol PCl3 137.33g PCl3


= 118g PCl3
70.91g Cl2 6 mol Cl 2
1mol PCl3
Thus, the limiting reactant is P4 , and 111 g PCl 3 should be produced. This is the theoretical
maximum yield. The actual yield is 104 g PCl 3 . Thus, the percent yield of the reaction is
104 g PCl 3 produced
 100 %  93.7% yield.
111g PCl 3 calculated
mass PCl3 = 91.5g Cl 2 
15A The reaction is 2 NH 3 (g) + CO 2 (g)  CO  NH 2 2 (s) + H 2 O(l) . We need to distinguish
between mass of urea produced (actual yield) and mass of urea predicted (theoretical yield).
mass CO 2 = 50.0 g CO  NH 2 2 produced 

44.01g CO 2
1mol CO 2
100.0 g predicted
87.5 g produced
 41.8 g CO 2 needed
71

1mol CO  NH 2 2
60.1g CO  NH 2 2

1mol CO 2
1mol CO  NH 2 2
Chapter 4: Chemical Reactions
15B Care must be taken to use the proper units/labels in each conversion factor. Note, you
cannot calculate the molar mass of an impure material or mixture.
mass C6 H11OH = 45.0 g C 6 H10 produced 

100.0 g C6 H10 cal'd

1mol C 6 H10
86.2 g C 6 H10 produc'd 82.1g C 6 H10

1mol C 6 H11OH
1mol C6 H10
100.2 g pure C6 H11OH 100.0 g impure C 6 H11OH

 69.0 g impure C 6 H11OH
1mol C 6 H11OH
92.3 g pure C 6 H11OH
16A We can trace the nitrogen through the sequence of reactions. We notice that 4 moles of N
(as 4 mol NH 3 ) are consumed in the first reaction, and 4 moles of N (as 4 mol NO) are
produced. In the second reaction, 2 moles of N (as 2 mol NO) are consumed and 2 moles of
N (as 2 mol NO 2 ) are produced. In the last reaction, 3 moles of N (as 3 mol NO 2 ) are
consumed and just 2 moles of N (as 2 mol HNO 3 ) are produced.
1000 g NH 3 1mol NH 3
4 mol NO 2 mol NO 2
mass HNO3 = 1.00 kg NH 3 



1kg NH 3
17.03g NH 3 4 mol NH 3 2 mol NO
2 mol HNO 3 63.01g HNO 3


= 2.47  103 g HNO 3
3 mol NO 2
1 mol HNO 3
16B
mass KNO3 = 95 g NaN 3 
1 mol NaN 3
2 mol KNO3
102 g KNO3
2 mol Na



65.03 g NaN3
2 mol NaN 3
10 mol Na
1 mol KNO3
 29.80  30 g KNO3
mass SiO 2 (1) = 1.461 mol NaN 3 
2 mol Na
1 mol K 2 O
1 mol SiO 2
64.06 g SiO 2



2 mol NaN 3
10 mol Na
1 mol K 2 O
1 mol SiO 2
 9.36 g  9.4 g SiO 2
mass SiO 2 (2) = 1.461 mol NaN 3 
2 mol Na
5 mol Na 2 O
1 mol SiO 2
64.06 g SiO 2



2 mol NaN 3
10 mol Na
1 mol Na 2 O
1 mol SiO 2
 46.80 g  47 g SiO 2
Therefore, the total mass of SiO2 is the sum of the above two results. Approximately 56 g
of SiO2 and 30 g of KNO3 are needed.
17A
To determine the mass% for each element,
1 mol Al 3 mol H 2 2.016 g H 2
mass Al = (m) g Al ×
×
×
= (m) 0.1121 g Al
26.98 g Al 2 mol Al 1 mol H 2
mass Mg = (1.00-m) g Al ×
1 mol Mg
×
1 mol H 2
24.305 g Mg 1 mol Mg
×
2.016 g H 2
1 mol H 2
= (1.00-m) 0.0829 g Mg
Now, we note that the total mass of H2 generated is 0.107 g. Therefore,
72
Chapter 4: Chemical Reactions
Mass H2 = (m)(0.1121) + (1.00-m)(0.0829) = 0.107
Solving for m gives a value of 0.82 g.
Therefore, mass of Al = 0.83 g. Since the sample is 1.00 g, Mg is 17 wt%..
mass of Mg = 1.00 – 0.83 = 0.17 g, or 17 wt%.
17B Mass of CuO and Cu2O is done in identical fashion to the above problem:
mass CuO = (1.500-x) g CuO ×
1 mol CuO
×
1 mol Cu
79.545 g CuO 1 mol CuO
×
63.546 g Cu
1 mol Cu
= (1.500-x) 0.7989
mass Cu 2 O = (x) g Cu 2 O ×
1 mol Cu 2 O
×
2 mol Cu
143.091 g Cu 2 O 1 mol Cu 2 O
×
63.546 g Cu
1 mol Cu
= (x) 0.8882 g Cu 2 O
Now, we note that the total mass of pure Cu is 1.2244 g. Therefore,
Mass Cu = (1.500-x)(0.7989) + (x)(0.8882) = 1.2244
Solving for x gives a value of 0.292 g.
Therefore, mass of Cu2O = 0.292
mass % of Cu2O = 0.292/1.500 × 100 = 19.47%
INTEGRATIVE EXAMPLE
A.
Balancing the equation gives the following:
C6 H10 O4 (l)  2NH 3 (g)  4H 2  C6 H16 N 2 (l)  4H 2 O
Stepwise approach:
The first step is to calculate the number of moles of each reactant from the masses given.
1000 g
1 mol.

 28.4 mol
1 kg
146.16 g
1000 g
1 mol.

 32.1 mol
mol NH3 = 0.547 kg 
1 kg
17.03 g
1000 g
1 mol.

 85.3 mol
mol H 2 = 0.172 kg 
1 kg
2.016 g
mol C6 H10 O 4 = 4.15 kg 
73
Chapter 4: Chemical Reactions
To determine the limiting reagent, calculate the number of moles of product that can be obtained
from each of the reactants. The reactant yielding the least amount of product is the limiting
reagent.
mol of C6 H16 N 2 from C6 H10 O4 = 28.4 mol 
mol of C6 H16 N 2 from NH3 = 32.1 mol 
1 mol C6 H16 N 2
 28.4 mol
1 mol C6 H10 O 4
1 mol C6 H16 N 2
 16.05 mol
2 mol NH3
1 mol C6 H16 N 2
 21.3 mol
4 mol H 2
NH3 yields the fewest moles of product, and is the limiting reagent.
mol of C6 H16 N 2 from H 2 = 85.3 
To calculate the % yield, the theoretical yield must first be calculated using the limiting reagent:
Theoretical yield = 16.05 mol C6 H16 N 2 
% yield =
116.22 g C6 H16 N 2
1 kg

 1.865 kg
1 mol C6 H16 N 2
1000 g
1.46 kg
 78.3% yield
1.865 kg
Conversion pathway Approach:
mol of C6 H16 N 2 from C6 H10 O4 = 4.15 kg C6 H10 O4 
mol of C6 H16 N 2 from NH3 = 0.547 kg NH3 
1 mol C6 H16 N 2
1000 g
1 mol.


 28.4 mol
1 kg
146.16 g
1 mol C6 H10 O 4
1 mol C6 H16 N 2
1000 g
1 mol.
 16.05 mol


1 kg
17.03 g
2 mol NH 3
1 mol C6 H16 N 2
1000 g
1 mol.


 21.3 mol
1 kg
2.016 g
4 mol H 2
NH3 yields the fewest moles of product and is therefore the limiting reagent
The % yield is determined exactly as above
mol of C6 H16 N 2 from H 2 = 0.172 kg H 2 
B.
Balancing the equation gives the following:
Zn(s) + 2HCl(aq)  ZnCl2 (aq) + H 2 (g)
Stepwise approach:
To determine the amount of zinc in sample, the amount of HCl reacted has to be calculated first:
Before reaction: 0.0179 M HCl  750.0 mL 
74
1L
 0.0134 mol HCl
1000 mL
Chapter 4: Chemical Reactions
After reaction: 0.0043 M HCl  750.0 mL 
1L
 0.00323 mol HCl
1000 mL
moles of HCl consumed = 0.0134 – 0.00323 = 0.0102 mol
Based on the number of moles of HCl consumed, the number of moles of Zn reacted can be
determined:
1 mol Zn
65.39 g Zn

 0.3335 g Zn
2 mol HCl
1 mol Zn
0.3335 g Zn reacted
Purity of Zn =
 100 = 83.4% pure
0.4000 g Zn in sample
0.0102 mol HCl 
Conversion pathway Approach:
 0.0179 - 0.0043 M HCl 
 750.0 mL 
1L
1000 mL
 0.0102 mol HCl
Note that we can only subtract concentrations in the above example because the volume has not
changed. Had there been a volume change, we would have to individually convert each
concentration to moles first.
1 mol Zn
65.39 g Zn 


 0.0102 mol HCl 
 / 0.4000 g  100  83.4 % Zn
2 mol HCl
1 mol Zn 

EXERCISES
Writing and Balancing Chemical Equations
1.
(a) 2 SO3 
 2 SO 2  O 2
(b) Cl2 O7  H 2 O 
 2 HClO 4
(c) 3 NO 2  H 2 O 
 2 HNO3  NO
(d) PCl3  3 H 2 O 
 H 3 PO3  3 HCl
3.
(a) 3 PbO + 2 NH 3 
 3 Pb + N 2 + 3 H 2 O
(b) 2 FeSO 4 
 Fe 2 O3  2 SO 2 +
1
2
O 2 or 4 FeSO 4 
 2 Fe 2 O3  4 SO 2 + O 2
(c) 6 S2 Cl2  16 NH 3 
 N 4 S4  12 NH 4 Cl +S8
(d) C3 H 7 CHOHCH(C2 H 5 )CH 2 OH  23 2 O 2 
 8 CO 2 + 9 H 2 O
or
2 C3 H 7 CHOHCH(C2 H5 )CH 2 OH  23 O 2 
16 CO 2 +18 H 2 O
75
Chapter 4: Chemical Reactions
5.
(a) 2 Mg  s  + O 2  g   2 MgO  s 
(b) 2 NO  g  + O 2  g   2 NO 2  g 
(c) 2 C2 H 6  g  + 7 O 2  g   4 CO 2  g  + 6 H 2 O  l 
(d) Ag 2 SO 4  aq  + BaI2  aq   BaSO4  s  + 2 AgI  s 
7.
(a) 2 C4 H10 (l) +13O 2  g   8CO 2  g  +10 H 2 O  l 
(b) 2 CH 3 CH(OH)CH 3 (l) + 9 O 2  g   6 CO 2  g  + 8 H 2 O  l 
(c) CH 3 CH(OH)COOH  s  + 3O 2  g   3CO 2  g  + 3H 2 O  l 
9.

(a) NH 4 NO3  s  
 N2 O g  + 2 H2 O g 
(b) Na 2 CO3  aq  + 2 HCl  aq   2 NaCl  aq  + H 2 O  l  + CO 2  g 
(c) 2 CH 4  g  + 2 NH 3  g  + 3O 2  g   2 HCN  g  + 6 H 2 O  g 
11.
Unbalanced reaction:
Balance H atoms:
Balance O atoms:
Balance N atoms:
Multiply by 2 (whole #)
Self Check:
N2H4(g) + N2O4(g)  H2O(g) + N2(g)
N2H4(g) + N2O4(g)  2 H2O(g) + N2(g)
N2H4(g) + 1/2 N2O4(g)  2 H2O(g) + N2(g)
N2H4(g) + 1/2 N2O4(g)  2 H2O(g) + 3/2 N2(g)
2 N2H4(g) + N2O4(g)  4 H2O(g) + 3 N2(g)
6N+8H+4O
 6N+8H+4O
Stoichiometry of Chemical Reactions
13.
In order to write the balanced chemical equation for the reaction, we will need to determine
the formula of the chromium oxide product.
First determine the number of moles of chromium and oxygen, and then calculate the mole
ratio.
1 mol Cr
 0.01325 mol Cr
52.00 g Cr
1 mol O 2
2 mol O
# mol O  0.636 g O 2 

 0.03975 mol O
32.00 g O 2 1 mol O 2
# mol Cr  0.689 g Cr 
0.03975 mol O
3 mol O
=
Therefore, the formula for the product is CrO3.
0.01325 mol Cr 1 mol Cr
Balanced equation =
2 Cr(s) + 3 O2(g)  2 CrO3(s)
76
Chapter 4: Chemical Reactions
15.
The conversion factor is obtained from the balanced chemical equation.
1mol Cl2
 7.26 mol Cl2
70.90 g Cl2
2 mol FeCl 3
= 4.84 mol FeCl 3
moles FeCl 3 = 7.26 mol Cl 2 
3 mol Cl 2
515 g Cl2 
17.
(a) Conversion pathway approach:
1mol KClO3
3mol O 2
32.8g KClO3 
= 0.401mol O 2

122.6 g KClO3
2 mol KClO3
Stepwise approach:
1mol KClO3
32.8g KClO3 
= 0.268 mol KClO3
122.6 g KClO3
0.268 mol KClO3 
(b)
3mol O 2
= 0.402 mol O 2
2 mol KClO3
Conversion pathway approach:
mass KClO3 = 50.0 g O 2 
1mol O 2
32.00 g O 2

2 mol KClO3 122.6 g KClO3
128g KClO 3

3 mol O 2
1mol KClO3
Stepwise approach:
50.0 g O 2 
1mol O 2
32.00 g O 2
1.56 mol O 2 
2 mol KClO3
3 mol O 2
1.04 mol KClO3 
(c)
= 1.56 mol O 2
= 1.04 mol KClO3
122.6 g KClO3
1mol KClO3
= 128 g KClO3
Conversion pathway approach:
mass KCl = 28.3g O 2 
1mol O 2 2 mol KCl 74.55g KCl


= 43.9 g KCl
32.00 g O 2 3mol O 2
1mol KCl
Stepwise approach:
77
Chapter 4: Chemical Reactions
28.3g O 2 
1mol O 2
= 0.884 mol O 2
32.00 g O 2
0.884 mol O 2 
2 mol KCl
= 0.589 mol KCl
3mol O 2
0.589 mol KCl 
19.
74.55g KCl
= 43.9 g KCl
1mol KCl
Balance the given equation, and then solve the problem.

2 Ag 2 CO3  s  
 4Ag  s  + 2 CO 2  g  + O 2  g 
mass Ag 2 CO 3 = 75.1g Ag 
21.
1mol Ag
107.87 g Ag

2 mol Ag 2 CO3
4 mol Ag

275.75 g Ag 2 CO3
1mol Ag 2 CO 3
= 96.0 g Ag 2 CO 3
The balanced equation is CaH2(s) + 2 H2O(l)  Ca(OH)2(s) + 2 H2(g)
1mol CaH 2
2 mol H 2
2.016 g H 2
(a) mass H 2 = 127 g CaH 2 


= 12.2 g H 2
42.094 g CaH 2 1mol CaH 2 1mol H 2
1mol CaH 2
2 mol H 2 O 18.0153 g H 2 O
(b) mass H 2 O = 56.2 g CaH 2 
= 48.1 g H 2 O


42.094 g CaH 2 1mol CaH 2
1 mol H 2 O
(c)
mass CaH 2 = 8.12×1024 molecules H 2 ×
1mol H 2
6.022×10
23
molecules H 2
×
1mol CaH 2
2 mol H 2
×
42.094 g CaH 2
1mol CaH 2
mass CaH 2 = 284 g CaH 2
23.

The balanced equation is Fe 2 O3  s  + 3C  s  
 2 Fe  l  + 3CO  g 
1 kmol Fe 1 kmol Fe 2 O 3 159.7 kg Fe 2 O 3


= 748 kg Fe 2 O 3
55.85 kg Fe
2 kmol Fe
1 kmol Fe 2 O 3
748 kg Fe 2 O 3
% Fe 2 O 3 in ore =
 100% = 79.7% Fe 2 O 3
938 kg ore
mass Fe 2 O 3 = 523 kg Fe 
25.
B10H14 + 11 O2  5 B2O3 + 7 H2O
% by mass B10 H14 =
# g B10 H14
 100
# g B10 H14 + # g O 2
1 mol B10H14 reacts with 11 mol O2 (exactly)
mass B10 H14 = 1 mol B10 H14 
mass O 2 = 11 mol O 2 
122.22 g B10 H14
= 122.22 g B10 H14
1 mol B10 H14
32.00 g O 2
= 352.00 g O 2
1 mol O 2
78
Chapter 4: Chemical Reactions
% by mass B10 H14 =
27.
122.22 g B10 H14
 100 = 25.8%
122.22 g B10 H14 + 352.00 g O 2
2 Al  s  + 6 HCl  aq   2 AlCl3  aq  + 3H 2  g  . First determine the mass of Al in the foil.
1cm 2.70 g

= 9.15g Al
10 mm 1cm3
1 mol Al 3 mol H 2 2.016 g H 2
mass H 2 = 9.15 g Al 


= 1.03 g H 2
26.98 g Al 2 mol Al 1 mol H 2
mass Al = 10.25cm  5.50 cm  0.601mm  
29.
First write the balanced chemical equation for each reaction.
2 Na  s  + 2 HCl  aq   2 NaCl  aq  + H 2  g 
2 Al  s  + 6 HCl  aq   2 AlCl3  aq  + 3 H 2  g 
Mg  s  + 2 HCl  aq   MgCl 2  aq  + H 2  g 
Zn  s  + 2 HCl  aq   ZnCl 2  aq  + H 2  g 
Three of the reactions—those of Na, Mg, and Zn—produce 1 mole of H2(g). The one of
these three that produces the most hydrogen per gram of metal is the one for which the
metal’s atomic mass is the smallest, remembering to compare twice the atomic mass for Na.
The atomic masses are: 2  23 u for Na, 24.3 u for Mg, and 65.4 u for Zn. Thus, among
these three, Mg produces the most H 2 per gram of metal, specifically 1 mol H 2 per 24.3 g
Mg. In the case of Al, 3 moles of H 2 are produced by 2 moles of the metal, or 54 g Al. This
reduces as follows: 3 mol H 2 / 54 g Al = 1 mol H 2 / 18 g Al. Thus, Al produces the largest
amount of H 2 per gram of metal.
Molarity
31.
(a)
(b)
(c)
2.92 mol CH 3 OH
= 0.408 M
7.16 L
7.69 mmol CH 3CH 2OH
CH 3CH 2OH molarity (M) =
= 0.154 M
50.00 mL
CH 3 OH molarity (M) =
CO  NH 2  2 molarity (M) =
25.2 g CO  NH 2  2
275 mL

1mol CO  NH 2 2
60.06 g CO  NH 2  2

1000 mL
1L
= 1.53 M
33.
(a)
Conversion pathway approach:
150.0 g C12 H 22 O11 1000 mL 1 mol C12 H 22 O11
[C12 H 22 O11 ] =
×
×
= 1.753 M
250.0 mL soln
1L
342.3 g C12 H 22 O11
79
Chapter 4: Chemical Reactions
Stepwise approach:
150.0 g C12 H 22 O11×
1 mol C12 H 22 O11
= 0.4382 mol C12 H 22 O11
342.3 g C12 H 22 O11
1L
= 0.2500 L
1000 mL
0.4382 mol C12 H 22 O11
[C12 H 22 O11 ] =
= 1.753 M
0.2500 L
250.0 mL soln×
(b)
Conversion pathway approach:
98.3 mg solid 97.9 mg CO(NH 2 ) 2 1 mmol CO(NH 2 ) 2
[CO(NH 2 ) 2 ] =
×
×
5.00 mL soln
100 mg solid
60.06 mg CO(NH 2 ) 2
= 0.320 M CO(NH 2 ) 2
Stepwise approach:
98.3 mg solid ×
97.9 mg CO(NH 2 ) 2
 96.2 mg CO(NH 2 ) 2
100 mg solid
96.2 mg CO(NH 2 ) 2 ×
1 g CO(NH 2 ) 2
= 0.0962 g CO(NH 2 ) 2
1000 mg CO(NH 2 ) 2
0.0962 g CO(NH 2 ) 2 ×
1 mol CO(NH 2 ) 2
= 1.60 10-3 mol CO(NH 2 ) 2
60.06 g CO(NH 2 ) 2
1L
= 0.00500 L
1000 mL
1.60 10-3 mol CO(NH 2 ) 2
[CO(NH 2 ) 2 ] =
= 0.320 M
0.00500 L
5.00 mL soln ×
(c)
Conversion pathway approach:
125.0 mL CH 3 OH 0.792 g 1 mol CH 3 OH
[CH 3 OH] =
×
×
= 0.206 M
15.0 L soln
1 mL 32.04 g CH 3 OH
Stepwise approach:
125.0 mL CH 3OH 0.792 g 1 mol CH 3OH
[CH 3OH] =
×
×
= 0.206 M
15.0 L soln
1 mL 32.04 g CH 3OH
0.792 g
= 99.0 g CH 3OH
1 mL
1 mol CH 3OH
99.0 g CH 3OH ×
= 3.09 mol CH 3OH
32.04 g CH 3OH
125.0 mL CH 3OH ×
[CH 3OH] =
3.09 mol CH 3OH
= 0.206 M
15.0 L soln
80
Chapter 4: Chemical Reactions
35.
(a)
mass C6 H12 O 6 = 75.0 mL soln 
(b)
VCH3OH = 2.25 L soln 
1L
1000 mL

0.350 mol C 6 H12 O 6
1L soln

180.16 g C6 H12 O 6
1mol C 6 H12 O 6
= 4.73g
0.485 mol 32.04 g CH 3OH 1mL


= 44.1 mL CH 3OH
1L
1mol CH 3OH 0.792 g
85 mg C6 H12 O6
1 mmol C6 H12 O6
1g
10 dL 1 mol C6 H12 O6




1 dL blood
1000 mg 1 L 180.16 g C6 H12 O6 110-3 mol C6 H12 O6
mmol C6 H12 O6
= 4.7
L
mol C6 H12 O6
(b) Molarity  4.7  103
L
37.
(a)
39.
First we determine each concentration in moles per liter and find the 0.500 M solution.
0.500 g KCl
1 mol KCl
1000 mL
(a) [KCl] =


= 6.71 M KCl
1 mL
74.551 g KCl
1L
36.0 g KCl
1 mol KCl
(b) [KCl] =

= 0.483 M KCl
1L
74.551 g KCl
7.46 mg KCl
1 g KCl
1 mol KCl
1000 mL
(c) [KCl] =



= 0.100 M KCl
1 mL
1000 mg KCl 74.551 g KCl
1L
373 g KCl
1 mol KCl
(d) [KCl] =

= 0.500 M KCl
10.00 L
74.551 g KCl
Solution (d) is a 0.500 M KCl solution.
41. We determine the molar concentration for the 46% by mass sucrose solution.
1 mol C12 H 22 O11
46 g C12 H 22 O11 ×
342.3 g C12 H 22 O11
[C12 H 22 O11 ] =
= 1.6 M
1 mL
1L
100 g soln×
×
1.21 g soln 1000 mL
The 46% by mass sucrose solution is the more concentrated.
2.05 mol KNO3
1L
= 0.0820 M
0.250 L diluted solution
0.01000 L conc'd soln 
43.
[KNO3 ] =
45. Both the diluted and concentrated solutions contain the same number of moles of K 2SO 4 . This
number is given in the numerator of the following expression.
0.198 mol K 2 SO 4
0.125 L 
1L
K 2 SO 4 molarity =
= 0.236 M K2SO4
0.105 L
81
Chapter 4: Chemical Reactions
47. Let us compute how many mL of dilute (d) solution we obtain from each mL of concentrated
(c) solution. Vc  Cc = Vd  Cd becomes 1.00 mL  0.250M = x mL  0.0125 M and x = 20
Thus, the ratio of the volume of the volumetric flask to that of the pipet would be 20:1. We
could use a 100.0-mL flask and a 5.00-mL pipet, a 1000.0-mL flask and a 50.00-mL pipet,
or a 500.0-mL flask and a 25.00-mL pipet. There are many combinations that could be
used.
Chemical Reactions in Solutions
49.
(a)
1L
0.163 mol AgNO 3
1 mol Na 2S


1000 mL
1 L soln
2 mol AgNO 3
78.05g Na 2S
= 0.177 g Na 2 S

1mol Na 2 S
mass Na 2S = 27.8 mL 
(b) mass Ag 2S = 0.177 g Na 2S 
51.
1mol Ag 2S 247.80 g Ag 2S

= 0.562 g Ag 2S
78.05 g Na 2S 1mol Na 2S
1mol Ag 2S

The molarity can be expressed as millimoles of solute per milliliter of solution.
0.186 mmol AgNO3 1mmol K 2 CrO 4 1mL K 2 CrO 4  aq 
VK CrO = 415 mL 


2
4
1mL soln
2 mmol AgNO3 0.650 mmol K 2 CrO 4
VK
53.
1mol Na 2S
2CrO4
= 59.4 mL K 2CrO 4
The balanced chemical equation for the reaction is:
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2H2O(l)
# mol HNO3  0.02978 L soln 
0.0142 mol Ca(OH) 2 2 mol HNO3

 8.46 104 mol HNO3
1 L soln
1 mol Ca(OH) 2
All of the HNO3 that reacts was contained in the initial, undiluted 1.00 mL sample. Since
the moles of HNO3 are the same in the diluted and undiluted solutions, one can divide the
moles of HNO3 by the volume of the undiluted solution to obtain the molarity.
Molarity 
55.
8.46 104 mol HNO3
mol HNO3
 8.46  103
0.00100 L
L
We know that the Al forms the AlCl 3 .
1mol Al 1mol AlCl3
mol AlCl3 = 1.87 g Al 
= 0.0693mol AlCl3

26.98g Al
1mol Al
0.0693 mol AlCl3 1000 mL
= 2.91 M AlCl3
(b) [AlCl3 ] =

23.8 mL
1L
(a)
82
Chapter 4: Chemical Reactions
57. The volume of solution determines the amount of product.
0.186 mol AgNO3 1mol Ag 2 CrO 4 331.73g Ag 2 CrO 4
1L
mass Ag 2 CrO 4 = 415 mL 



1000 mL
1L soln
2 mol AgNO3
1mol Ag 2 CrO 4
mass Ag 2 CrO 4 = 12.8 g Ag 2 CrO 4
59.
mass Na = 155mL soln 
1L
0.175mol NaOH
2 mol Na
22.99g Na



1000 mL
1L soln
2 mol NaOH 1mol Na
= 0.624g Na
61. The mass of oxalic acid enables us to determine the amount of NaOH in the solution.
0.3126 g H 2 C 2O 4 1000 mL 1mol H 2C 2 O 4
2 mol NaOH



= 0.2649 M
 NaOH  =
26.21mL soln
1L soln 90.04 g H 2 C2 O 4 1mol H 2 C2 O 4
Determining the Limiting Reactant
63. The limiting reactant is NH 3 . For every mole of NH3(g) that reacts, a mole of NO(g) forms.
Since 3.00 moles of NH3(g) reacts, 3.00 moles of NO(g) forms (1:1 mole ratio).
65. First we must determine the number of moles of NO produced by each reactant. The one
producing the smaller amount of NO is the limiting reactant.
2 mol NO
mol NO = 0.696 mol Cu 
= 0.464 mol NO
3mol Cu
Conversion pathway approach:
mol NO = 136 mL HNO3  aq  ×
6.0 mol HNO3
1L
2 mol NO
×
×
= 0.204 mol NO
1000 mL
1L
8 mol HNO3
Stepwise approach:
1L
 0.136 L HNO3
1000 mL
6.0 mol HNO3
0.136 L ×
 0.816 mol HNO3
1L
2 mol NO
0.816 mol HNO3×
= 0.204 mol NO
8 mol HNO3
136 mL HNO3  aq  ×
Since HNO3(aq) is the limiting reactant, it will be completely consumed, leaving some Cu
unreacted.
83
Chapter 4: Chemical Reactions
67. First we need to determine the amount of Na 2 CS3 produced from each of the reactants.
2 mol Na 2 CS3
1.26 g 1mol CS2
n Na 2CS3 (from CS2 ) = 92.5 mL CS2 
= 1.02 mol Na 2 CS3


1mL 76.14 g CS2
3mol CS2
2 mol Na 2 CS3
n Na 2CS3 (from NaOH) = 2.78 mol NaOH 
= 0.927 mol Na 2 CS3
6 mol NaOH
154.2 g Na 2 CS3
= 143g Na 2 CS3
Thus, the mass produced is 0.927 mol Na 2 CS3 
1mol Na 2 CS3
69.
Ca  OH 2  s  + 2 NH 4 Cl  s   CaCl2  aq  + 2 H 2 O(l) + 2 NH 3  g 
First compute the amount of NH 3 formed from each reactant in this limiting reactant problem.
2 mol NH 3
1mol NH 4 Cl
n NH3 (from NH 4 Cl) = 33.0 g NH 4 Cl 
= 0.617 mol NH 3

53.49 g NH 4 Cl 2 mol NH 4 Cl
1mol Ca  OH 2
2 mol NH 3
= 0.891mol NH 3
74.09 g Ca  OH 2 1mol Ca  OH 2
Thus, 0.617 mol NH 3 should be produced as NH4Cl is the limiting reagent.
n NH3 (from Ca(OH) 2 ) = 33.0 g Ca  OH 2 
mass NH 3 = 0.617 mol NH 3 
17.03g NH 3
1mol NH 3

= 10.5 g NH 3
Now we will determine the mass of reactant in excess, Ca(OH)2.
1mol Ca  OH 2 74.09 g Ca  OH 2
Ca  OH 2 used = 0.617 mol NH3 

= 22.9 g Ca OH
2 mol NH3
1mol Ca  OH 2
b g
2
excess mass Ca  OH 2 = 33.0 g Ca  OH 2  22.9 g Ca  OH 2 = 10.1 g excess Ca  OH 2
71. The number of grams of CrSO4 that can be made from the reaction mixture is determined by
finding the limiting reagent, and using the limiting reagent to calculate the mass of product
that can be formed. The limiting reagent can determined by calculating the amount of
product formed from each of the reactants. Whichever reactant produces the smallest
amount of product is the limiting reagent.
2 mol CrSO 4 148.06 g CrSO 4

= 236.90 g CrSO 4
4 mol Zn
1 mol CrSO 4
2 mol CrSO 4
148.06 g CrSO 4
1.7 mol K 2Cr2 O7 

= 503.40 g CrSO 4
1 mol K 2 Cr2 O7
1 mol CrSO 4
3.2 mol Zn 
5.0 mol H 2SO 4 
2 mol CrSO 4 148.06 g CrSO 4

= 211.51 g CrSO 4
7 mol H 2SO 4
1 mol CrSO 4
H2SO4 is the limiting reagent since it produces the least amount of CrSO4. Therefore, the
maximum number of grams of CrSO4 that can be made is 211.51 g.
84
Chapter 4: Chemical Reactions
Theoretical, Actual, and Percent Yields
73.
75.
1 mol CCl4
= 1.80 mol CCl4
153.81g CCl4
Since the stoichiometry indicates that 1 mole CCl 2 F2 is produced per mole CCl 4 , the
use of 1.80 mole CCl 4 should produce 1.80 mole CCl2 F2 . This is the theoretical yield
of the reaction.
1mol CCl2 F2
(b) 187g CCl2 F2 
= 1.55 mol CCl2 F2
120.91g CCl2 F2
The actual yield of the reaction is the amount actually produced, 1.55 mol CCl2 F2 .
1.55mol CCl 2 F2 obtained
(c) % yield =
 100 %  861%
. yield
1.80 mol CCl 2 F2 calculated
(a)
277 grams CCl 4 
actual yield
100 %
theoretical yield
The actual yield is given in the problem and is equal to 28.2 g.
In order to determine the theoretical yield, we must find the limiting reagent and do
stoichiometry.
% yield =
Conversion pathway approach:
1 mol Al 2 O3
2 mol Na 3 AlF6 209.94 g Na 3 AlF6
7.81 g Al 2 O3 


 32.2 g Na 3 AlF6
101.96 g Al 2 O3 1 mol Al 2 O3
1 mol Na 3 AlF6
0.141 mol 2 mol Na 3 AlF6 209.94 g Na 3 AlF6
3.50 L 


 34.5 g Na 3 AlF6
1L
6 mol NaOH
1 mol Na 3 AlF6
Stepwise approach:
Amount of Na3AlF6 produced from Al2O3 if all Al2O3 reacts
1 mol Al 2 O3
7.81 g Al 2 O3 
= 0.0766 mol Al 2 O3
101.96 g Al 2 O3
0.0766 mol Al 2 O3 
2 mol Na 3 AlF6
= 0.153 mol Na 3 AlF6
1 mol Al 2 O3
209.94 g Na 3 AlF6
 32.1 g Na 3AlF6
1 mol Na 3 AlF6
Amount of Na3AlF6 produced from NaOH if all NaOH reacts
0.153 mol Na 3AlF6 
85
Chapter 4: Chemical Reactions
3.50 L 
0.141 mol NaOH
 0.494 mol NaOH
1L
2 mol Na 3 AlF6
= 0.165 mol Na 3 AlF6
6 mol NaOH
209.94 g Na 3 AlF6
0.165 mol Na 3 AlF6 
 34.5 g Na 3AlF6
1 mol Na 3 AlF6
0.494 mol NaOH 
Al2O3 is the limiting reagent.
% yield =
28.2 g
100 %  87.6%
32.2 g
81. A less-than-100% yield of desired product in synthesis reactions is always the case. This is because
of side reactions that yield products other than those desired and because of the loss of material on
the glassware, on filter paper, etc. during the various steps of the procedure. A main criterion for
choosing a synthesis reaction is how economically it can be run. In the analysis of a compound, on
the other hand, it is essential that all of the material present be detected. Therefore, a 100% yield is
required; none of the material present in the sample can be lost during the analysis. Therefore
analysis reactions are carefully chosen to meet this 100 % yield criterion; they need not be
economical to run.
Consecutive Reactions, Simultaneous Reactions
84.
Here we need to determine the amount of CO 2 produced from each reactant.
C3 H8  g  + 5O 2  g  
 3CO 2  g  + 4 H 2 O  l 
2 C4 H10  g  +13O 2  g  
 8CO 2  g  +10 H 2 O  l 
72.7g C3 H8
1mol C3 H8
3mol CO 2


= 20.1mol CO 2
100.0 g mixt. 44.10 g C3 H8 1mol C3 H8
27.3g C4 H10 1mol C4 H10
8 mol CO 2


n CO2 (from C 4 H10 ) = 406 g mixt. 
= 7.63mol CO 2
100.0 g mixt 58.12 g C4 H10 2 mol C4 H10
44.01g CO 2
mass CO 2 =  20.1 + 7.63 mol CO 2 
= 1.22  103 g CO 2
1mol CO 2
n CO2 (from C3 H 8 ) = 406 g mixt. 
86. Balanced Equations:
C2H6(g) + 7 2 O2(g)  2 CO2(g) + 3 H2O(l)
CO2(g) + Ba(OH)2(aq)  BaCO3(s) + H2O(l)
Conversion pathway approach:
mass C2 H6 = 0.506 g BaCO3 
1mol BaCO3
197.3g BaCO3

Stepwise approach:
86
1mol CO 2
1mol BaCO3

2 mol C 2 H 6
4 mol CO 2

30.07 g C 2 H 6
1mol C 2 H 6
= 0.0386 g C 2 H 6
Chapter 4: Chemical Reactions
0.506 g BaCO 3 
1mol BaCO3
197.3g BaCO3
2.56  10-3mol BaCO3 
2.56  10-3mol CO 2 
= 2.56  10-3 mol BaCO3
1mol CO 2
1mol BaCO3
2 mol C 2 H 6
4 mol CO 2
1.28  10-3 mol C 2 H 6 
= 2.56 10-3 mol CO 2
= 1.28  10-3 mol C2 H 6
30.07 g C 2 H 6
= 0.0386 g C 2 H 6
1 mol C 2 H 6
87. NaI(aq)+ AgNO3(aq)  AgI(s )+ NaNO3(aq)
(multiply by 4)
(multiply by 2)
2 AgI(s) + Fe(s)
 FeI2(aq) + 2 Ag(s)
(unchanged)
2 FeI2(aq) + 3 Cl2(g)  2 FeCl3(aq) + 2 I2(s)
4NaI(aq) + 4AgNO3(aq) + 2Fe(s) + 3Cl2(g) 4NaNO3(aq) + 4Ag(s) + 2FeCl3(aq) + 2I2(s)
For every 4 moles of AgNO3, 2 moles of I2(s) are produced. The mass of AgNO3 required
1 mol I 2 (s)
4 mol AgNO3 (s) 169.873 g AgNO3 (s)
1000 g I 2 (s)



= 1.00 kg I2(s)
253.809 g I 2 (s)
2 mol I 2 (s)
1 mol AgNO3 (s)
1 kg I 2 (s)
= 1338.59 g AgNO3 per kg of I2 produced or 1.34 kg AgNO3 per kg of I2 produced
89.
(a)

SiO2(s) + 2 C(s) 
 Si(s) + 2 CO(g)
Si(s) + 2 Cl2(g)
→ SiCl4(l)
SiCl4(l) + 2 H2(g) → Si(s, ultrapure) + 4 HCl(g)
(b)
1 kg Si (ultrapure, s) 
1 mol SiCl 4
1000 g
1 mol Si
1 mol Si



1kg
28.09 g
1 mol Si ultrapure
1 mol SiCl4
2 mol C
12.01 g C

= 885 g C
1 mol Si
1 mol C
1 mol SiCl 4
2 mol Cl2
1000 g
1 mol Si
1 kg Si (ultrapure, s) 



1kg
28.09 g
1 mol Si ultrapure
1 mol SiCl4


70.91 g Cl2
= 5.05  103 g Cl 2
1 mol Cl2
1 kg Si (ultrapure, s) 
1000 g
1 mol Si
2 mol H 2
2.016 g H 2



= 144 g H 2
1kg
28.09 g
1 mol Si ultrapure
1 mol H 2
87
Chapter 4: Chemical Reactions
91. The reactions are as follows.
MgCO3(s) → MgO(s) + CO2(g)
CaCO3(s) → CaO(s) + CO2(g)
% by mass of MgCO3 =
g MgCO3
 100 %
g MgCO3 + g CaCO3
Let m = mass, in grams, of MgCO3 in the mixture and
let 24.00 − m = mass in grams of CaCO3 in the mixture.
Convert from g MgCO3 to g CO2 to obtain an expression for the mass of CO2 produced by
the first reaction.
1 mol MgCO3
1 mol CO 2
44.01 g CO 2
g CO 2 from MgCO3  m g MgCO3 


84.32 g MgCO3 1 mol MgCO3 1 mol CO 2
Convert from g CaCO3 to g CO2 to obtain an expression for the mass of CO2 produced by
the second reaction.
1 mol CaCO3
1 mol CO 2
44.01 g CO 2
g CO 2 from CaCO3  (24.00  m) g CaCO3 


100.09 g CaCO3 1 mol CaCO3 1 mol CO 2
The sum of these two expressions is equal to 12.00 g CO2. Thus:
44.01 
44.01 


 m  84.32   (24.00  m)  100.09   12.00
Solve for m: m = 17.60 g
% by mass of MgCO3 =
17.60 g
 100 % = 73.33 %
24.00 g
INTEGRATIVE AND ADVANCED EXERCISES

93. (a) CaCO 3 (s) 

CaO(s)  CO 2 (g)

(b) 2 ZnS(s)  3 O 2 (g) 

2 ZnO(s)  2 SO 2 (g)
(c) C 3 H 8 (g)  3 H 2 O(g) 
 3 CO(g)  7 H 2 (g)
(d) 4 SO2(g) + 2 Na2S(aq) + Na2CO3(aq) 
 CO2(g) + 3 Na2S2O3(aq)
88
Chapter 4: Chemical Reactions
95. The balanced equation is as follows:
2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Conversion pathway approach:
g LiOH 
1.00 103 g CO 2
1 mol CO 2
2 mol LiOH 23.95 g LiOH

 3 astronauts  6 days 

astronaut day
44.01 g CO2
1 mol CO2
1 mol LiOH
 1.96  104 g LiOH
Stepwise approach:
1.00  103 g CO 2
1 mol CO 2
mol CO 2
= 22.7

astronaut day
44.01 g CO 2
astronaut day
mol CO 2
mol CO 2
22.7
 3 astronauts  68.2
astronaut day
day
mol CO 2
68.2
 6 days  409 mol CO 2
day
2 mol LiOH
409 mol CO 2 
= 818 mol LiOH
1 mol CO 2
23.95 g LiOH
818 mol LiOH 
 1.96  104 g LiOH
1 mol LiOH
96.
mass CaCO3 = 0.981g CO 2 
% CaCO 3 =
1mol CO 2
44.01g CO 2

1mol CaCO 3 100.1g CaCO3

= 2.23g CaCO3
1mol CO 2
1mol CaCO3
2.23 g CaCO 3
 100% = 68.0% CaCO 3 (by mass)
3.28 g sample
98. Assume 100g of the compound FexSy, then:
Number of moles of S atoms = 36.5g/32.066 g S/mol = 1.138 moles
Number of moles of Fe atoms = 63.5g/ 55.847g Fe/mol = 1.137 moles
So the empirical formula for the iron –containing reactant is FeS
Assume 100g of the compound FexOy, then:
Number of moles of O atoms = 27.6g/16.0 g O/mol = 1.725 moles
Number of moles of Fe atoms = 72.4g/ 55.847g Fe/mol = 1.296 moles
So the empirical formula for the iron-containing product is Fe3O4
Balanced equation: 3 FeS + 5 O2  Fe3O4 + 3 SO2
89
Chapter 4: Chemical Reactions
99.
M CH 3CH 2OH =
mol CH 3CH 2 OH
volume of solution
mol CH 3CH 2OH  50.0 mL  0.7893
Molarity 
101.
g CH 3CH 2 OH 1 mol CH 3CH 2 OH

= 0.857 mol
mL
46.07 g CH 3CH 2 OH
0.857 mol CH 3CH 2OH
 8.88 M CH 3CH 2 OH
0.0965 L solution
Let V be the volume of 0.149 M HCl(aq) that is required.
moles of HCl in solution C = moles HCl in solution A + moles HCl in solution B
(V + 0.100) × 0.205 M =
(V × 0.149 M ) + (0.100 × 0.285 M)
Solve for V:
V = 0.143 L = 143 mL
105. We can compute the volume of Al that reacts with the given quantity of HCl.
1L
12.0 mol HCl 2 mol Al 27.0 g Al
1cm3
VAl  0.05 mL 




 0.002 cm3
1000 mL
1L
6 mol HCl 1mol Al 2.70 g Al
0.002 cm3 10 mm
volume
area 


 0.2 cm 2
thickness
0.10 mm
1cm
106.
Here we need to determine the amount of HCl before and after reaction; the difference is
the amount of HCl that reacted.
1.035 mol HCl
initial amount HCl  0.05000 L 
 0.05175 mol HCl
1L
0.812 mol HCl
final amount HCl  0.05000 L 
 0.0406 mol HCl
1L
1 mol Zn 65.39 g Zn
mass Zn  (0.05175  0.0406) mol HCl 

 0.365 g Zn
2 mol HCl 1 mol Zn
 2 CO 2 (g)  3 H 2 O(l)
114. CH 3CH 2OH(l)  3O 2 (g) 
(CH 3CH 2 ) 2 O (l)  6 O 2 (g) 
 4 CO 2 (g)  5 H 2O(l)
Since this is classic mixture problem, we can use the systems of equations method to find the mass
percents. First we let x be the mass of (C2H5)2O and y be the mass of CH3CH2OH. Thus,
x + y = 1.005 g or y = 1.005 g – x
We then construct a second equation involving x that relates the mass of carbon dioxide formed to
the masses of ethanol and diethyl ether., viz.
90
Chapter 4: Chemical Reactions
1.963 g CO 2 
1 mol  C2 H 5 2 O
1 mol CO 2
4 mol CO 2
 x g (C2 H 5 ) 2 O 

44.010 g CO 2
74.123 g  C2 H 5 2 O 1 mol  C2 H 5 2 O
 (1.005  x) g CH3CH 2 OH 
0.04460  0.05396 x  0.04363  0.04341x
1 mol CH 3CH 2 OH
2 mol CO 2

46.07 g CH3CH 2 OH 1 mol CH3CH 2 OH
x 
0.092 g  C2 H 5 2 O
0.04460  0.04363
 0.092 g  C2 H 5 2 O
0.05396  0.04341
 100%  9.2%  C2 H 5 2 O
1.005 g mixture
% CH 3CH 2 OH (by mass)  100.0% – 9.2% (C2 H5 )2 O  90.8% CH 3CH 2 OH
% (CH3CH 2 )2 O (by mass) 
115. % Cu (by mass) 
# g Cu
 100
0.7391 g mixture
Let x = the mass, in grams, of CuCl2 in the mixture.
Let 0.7391 – x = mass in grams of FeCl3.
Total moles AgNO3 = mol AgNO3 react with CuCl2 + mol AgNO3 react with FeCl3
Total moles AgNO3 = 0.8691 L 
0.1463 mol
= 0.01271 mol AgNO3
1L
To obtain an expression for the amount of AgNO3 consumed by the first reaction, convert from
grams of CuCl2 to moles of AgCl:
2 mol AgNO3
1 mol CuCl2
mol AgNO3 that reacts with CuCl2  x g CuCl2 

134.45 g CuCl2 1 mol CuCl2
2x
mol AgNO3 that reacts with CuCl 2 
= 0.014875x
134.45
To obtain an expression for the amount of AgNO3 consumed by the second reaction, convert from
grams of FeCl3 to moles of AgNO3:
1 mol FeCl3
3 mol AgNO3

162.21 g FeCl3 1 mol FeCl3
mol AgNO3 that reacts with FeCl3  (0.7391  x)  0.018496 = 0.013668  0.018496x
The sum of these two expressions is equal to the total number of moles of AgNO3 :
mol AgNO3 that reacts with FeCl3  (0.7391  x) g FeCl3 
Total moles AgNO3 = 0.014875x + 0.013668 – 0.018496x = 0.01271
x = 0.2646 g CuCl2
This is the mass of CuCl2 in the mixture. We must now convert this to the mass of Cu in the
mixture.
91
Chapter 4: Chemical Reactions
# g Cu  0.2646 g CuCl2 
% Cu 
1 mol CuCl2
1 mol Cu
63.546 g Cu


 0.1253 g Cu
134.45 g CuCl2 1 mol CuCl2
1 mol Cu
0.1253 g Cu
 100 %  16.95 %
0.7391 g
116.
1 mol Cu 2
 0.766 mol Cu 2  0.307 
 2.50 mol Cu 2
63.55 g Cu 2
222- 1 mol CrO 4
mol CrO 4 =35.6 g CrO 4 ×
= 0.307 mol CrO 4 2- ÷0.307 
1.00 mol CrO 4 2115.99 g
(a) mol Cu 2  48.7 g Cu 2 
1 mol OH mol OH =15.7 g OH ×
= 0.923 mol OH - ÷0.307 
 3.01 mol OH 17.01 g OH
Empirical formula: Cu 5 (CrO 4 ) 2 (OH)6
-
-
(b) 5 CuSO 4 (aq)  2 K 2 CrO 4 (aq)  6 H 2O (l)

Cu 5 (CrO 4 ) 2 (OH)6 (s)  2 K 2SO 4 (aq)  3 H 2SO 4 (aq)
117. We first need to compute the empirical formula of malonic acid.
1 mol C
34.62 g C 
 2.883 mol C
2.883 
 1.000 mol C
12.01 g C
1 mol O
3.88 g H 
 3.84 mol O
2.883 
 1.33 mol H
1.01 g H
1 mol O
61.50 g O 
 3.844 mol O
2.883 
 1.333 mol O
16.00 g O
Multiply each of these mole numbers by 3 to obtain the empirical formula C3H4O4.
Combustion reaction: C3 H 4 O 4 (l)  2 O 2 (g) 
 3 CO 2 (g)  2 H 2 O(l)
118.
2 Al (s) + Fe2O3 → Al2O3 + 2 Fe
1 mol Fe 2 O3
159.69 g Fe 2 O3
1 mol Al
mass of Fe 2 O3  2.5 g Al 


 7.4 g Fe 2O3 needed
26.982 g Al
2 mol Al
2 mol Fe 2 O3
Using 2.5 g Al2O3, only 7.4 g of Fe2O3 needed, but there are 9.5 g available. Therefore, Al
is the limiting reagent.
1 mol Al
2 mol Fe
55.85 g Fe


 5.2 g Fe
26.982 g Al
2 mol Al
1 mol Fe 2 O3
(b) Mass of excess Fe2O3 = 9.5 g – 7.4 = 2.1 g
(a) Mass of Fe  2.5 g Al 
92
Chapter 4: Chemical Reactions
119. Compute the amount of AgNO3 in the solution on hand and the amount of AgNO3 in the
desired solution. the difference is the amount of AgNO3 that must be added; simply convert
this amount to a mass.
0.0500 mmol AgNO 3
amount AgNO 3 present  50.00 mL 
 2.50 mmol AgNO 3
1 mL soln
0.0750 mmol AgNO 3
amount AgNO 3 desired  100.0 mL 
 7.50 mmol AgNO 3
1 mL soln
1 mol AgNO 3
169.9 g Ag NO 3
mass AgNO 3  (7.50  2.50) mmol AgNO 3 

1000 mmol AgNO 3
1 mol AgNO 3
 0.850 g AgNO 3
120. The balanced equation for the reaction is: S8(s) + 4 Cl2(g)  4 S2Cl2(l)
Both “a” and “b” are consistent with the stoichiometry of this equation. Neither bottom row
box is valid. Box (c) does not account for all the S8, since we started out with 3 molecules,
but end up with 1 S8 molecule and 4 S2Cl2 molecules. Box (d) shows a yield of 2 S8
molecules and 8 S2Cl2 molecules so we ended up with more sulfur atoms than we started
with. This, of course, violates the Law of Conservation of Mass.
121. The pertinent equations are as follows:

C3 N 3  OH 3 
 3 HNCO (g)

8 HNCO + 6 NO 2 
 7 N 2 + 8 CO 2 + 4 H 2 O
The above mole ratios are used to calculate moles of C3N3(OH)3 assuming 1.00 g of NO2.
1 mol C3 N 3 (OH)3
1 mol NO 2
8 mol HNCO
mass C3 N 3 (OH)3  1.00 g NO 2 


46.00 g NO 2
6 mol NO 2
3 mol HNCO

129.1 g mol C3 N 3 (OH)3
 1.25 g C3 N 3 (OH)3
1 mol C3 N 3 (OH)3
123. There are many ways one can go about answering this question. We must use all of the most
concentrated solution and dilute this solution down using the next most concentrated solution.
Hence, start with 345 mL of 01.29 M then add x mL of the 0.775 M solution. The value of x is
obtained by solving the following equation.
1.29 M  0.345 L   0.775 M  x 
1.25 M =
(0.345  x) L
1.25 M  (0.345  x) L = 1.29 M  0.345 L   0.775 M  x 
043125 + 1.25x  0.44505 0.775 x Thus, 0.0138  0.475 x
x  0.029 L or 29 mL
A total of (29 mL + 345 mL) = 374 mL may be prepared this way.
93
Chapter 4: Chemical Reactions
127.
(a) 2 C3H6(g) + 2 NH3(g) + 3 O2(g)  2 C3H3N(l) + 6 H2O(l)
(b) For every kilogram of propylene we get 0.73 kilogram of acrylonitrile; we can also say that
for every gram of propylene we get 0.73 gram of acrylonitrile. One gram of propylene is
0.0238 mol of propylene. The corresponding quantity of NH3 is 0.0238 mol or 0.405 g; then
because NH3 and C3H6 are required in the same molar amount (2:2) for the reaction, 0.405 of a
kg of NH3 will be required for every 0.73 of a kg of acrylonitrile. To get 1000 kg of
acrylonitrile we need, by simple proportion, 1000×(0.405)/0.73 = 555 kg NH3.
SELF-ASSESSMENT EXERCISES
135. The answer is (d). Start balancing in the following order: N, O, H and Cu
3 Cu (s) + 8 HNO3  3 Cu(NO3 ) 2 + 4 H 2 O + 2 NO
136. The answer is (d). To determine the number of moles of NH3, used the balanced equation:
2 mol NH 3
# moles NH 3 = 1 mol H 2 O 
= 0.666
3 mol H 2 O
137. The answer is (a). To determine the number of moles of NH3, use the balanced equation:
2 KMnO4 (s) + 10 KI + 8 H2SO4  6 K 2SO4 + 2 MnSO4 + 5 I2 + 8 H 2O
5 KMnO 4 
6 mol K 2SO 4
= 15 mol K 2SO 4
2 mol KMnO 4
6 mol K 2SO 4
= 3 mol K 2SO 4
10 mol KI
6 mol K 2SO 4
5 H 2SO 4 
= 3.75 mol K 2SO 4
8 mol H 2SO 4
5 KI 
138. The answer is (a). To determine the answer, used the balanced equation:
2 Ag 2 (CO3 ) (s)  4 Ag + 2 CO 2 + O 2
The ratio between O2 and CO2 is 1:2.
139. The answer is (c). To solve this, calculate the number of moles of NaNO3.
mol NaNO3 = 1.00 M × 1.00 L = 1.00 mol.
85.0 g
1.00 mol NaNO3 
= 85.0 g NaNO3
1 mol
Concentration = 85.0 g NaNO3/L. While (b) also technically gives you the correct value at
25 °C, it is not the definition of molarity.
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Chapter 4: Chemical Reactions
140. The answer is (d). There is no need for calculation, because a starting solution of 0.4 M is
needed to make a 0.50 M solution, and the only way to make a more concentrated solution
is to evaporate off some of the water.
141. The answer is (b). To determine the molarity, number of moles of LiBr need to be
determined first. Therefore, weight% concentration needs to be converted to number of
moles with the aid of the density:
Conc. = 5.30% by mass = 5.30 g LiBr/100 g solution
Volume of solution = mass / Density = 100 g sol'n 
1 mL
= 96.15 mL
1.040g
1 mol LiBr
= 0.0610 mol
86.84 g LiBr
0.0610 mol
1000 mL
Molarity =

= 0.635 M
96.15 mL
1L
mol LiBr = 5.30 g LiBr 
142. The answer is (d). To determine % yield, calculate the theoretical mole yield:
1 mol CCl2 F
mol CCl2 F = 2.00 mol CCl 4 
= 2.00 mol CCl2 F
1 mol CCl4
% yield =
1.70 mol
 100 = 85.0%
2.00 mol
143. To balance the below equations, balance C first, then O and finally H.
(a) 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
(b) For this part, we note that 25% of the available carbon atoms in C8H18 form CO and the
remainder for CO2. Therefore,
2 C8H18 + 25 O2 → 12 CO2 + 4 CO + 18 H2O
144. To determine the compound, the number of moles of each compound needs to be
determined, which then helps determine number of moles of emitted CO2:
95
Chapter 4: Chemical Reactions
mass CO 2 = 1.000 g CaCO3 
1 mol CaCO3
1 mol CO 2
44.0 g CO 2


100.08 g CaCO3
1 mol CaCO3
1 mol CO 2
= 0.4396 g CO 2
mass CO 2 = 1.000 g MgCO3 
1 mol MgCO3
1 mol CO 2
44.0 g CO 2


1 mol CO 2
84.30 g MgCO3
1 mol CaCO3
= 0.5219 g CO 2
mass CO 2 = 1.000 g CaCO3 MgCO3 

2 mol CO 2
1 mol dolomite

184.38 g dolomite
1 mol dolomite
44.0 g CO 2
= 0.4773 g CO 2
1 mol CO 2
Therefore, dolomite is the compound.
145. The answer is (b). First, the total amount of carbon in our mixture of CH4 and C2H6 must
be determined by using the amount of CO2
1 mol CO 2
1 mol C
12.01 g C


= 0.758 g C
44.01 g CO2
1 mol CO 2
1 mol C
Then, the amounts of CH4 and C2H6 can be determined by making sure that the moles of
carbon for both add up to 0.0631:
mass of C = 2.776 g CO 2 
 1 mol CH 4
1 mol C
12.01 g C 


xg  

1 mol CH 4
1 mol C 
 16.05 g CH 4
 1 mol C2 H 6
2 mol C
12.01 g C 


+ (1.000-x)  
 = 0.757 g C (from CO 2 )
1 mol C 2 H 6
1 mol C 
 30.08 g C2 H 6
0.748 x + (1-x)(0.798) = 0.757
x = mass of CH 4 = 0.82 g, or 82% of a 1.00 g sample
146. The answer is (c). To do this, perform a stepwise conversion of moles of reactants to moles
of products, as shown below:
4.00 mol NH 3 
2 mol HNO3
2 mol NO 2
4 mol NO


= 2.67 mol HNO3
4 mol NH 3
2 mol NO
3 mol NO 2
96