Lesson C10 Homework Solutions
26. Define the problem: Given the volume of a sample of air at STP and the volume fraction of CO in
ppm, determine the moles of CO.
Develop a plan: Use the volume fraction in terms of liters as a conversion factor to determine the liters
of CO. Use the molar volume of a gas at STP as a conversion factor to determine the moles of CO.
Execute the plan:
1.0 L air ×
950 L CO
1 mol CO
×
= 4.2 ×10−5 mol CO
1,000, 000 L air 22.414 L CO
Check your answer: Air has a very small proportion of CO. This small sample of air has a very small
amount of CO. This number makes sense.
28. Define the problem: Given the volume and pressure of a sample of gas in one flask and the volume of
a flask it is transferred to at the same temperature, determine the new pressure.
Develop a plan: Use Boyle’s law to relate volume to pressure.
P1V1 = P2V2
(unchanging T and n)
Execute the plan:
P V (75.0 mmHg) × (256 mL)
= 154 mmHg
P2 = 1 1 =
V2
(125 mL)
Check your answer: The smaller volume should have a larger pressure. This answer makes sense.
30. Use Boyle’s law to relate volume to pressure, as described in the answer to Question 27.
P V (735 mmHg) × (3.50 L)
= 172 mmHg
P2 = 1 1 =
V2
(15.0 L)
32. Define the problem: Given the original volume and temperature of a sample of gas in a syringe
(presumably at atmospheric pressure) and the new temperature of the sample, determine the new
volume (presumably still at atmospheric pressure).
Develop a plan: Convert the temperatures to Kelvin. Use Charles’ law to relate volume to absolute
temperature.
V1 V2
=
(P and n constant)
T1 T2
Execute the plan: (Assume the T1 reading has the same precision as the reading for T2.)
T1 = 20. °C + 273 = 293 K
T2 = 37 °C + 273 = 310. K
T
(310. K)
= 26.5 mL
V2 = V1 × 2 = (25.0 mL) ×
T1
(293 K)
Check your answer: Gas at higher temperature should have a larger volume.
36. Define the problem: Given the original pressure and temperature of gas in a tire, the assumption that
volume is unchanged, and the new temperature, determine the new pressure exerted by the gas in the
tire.
Develop a plan: Convert the temperature to Kelvin. Use the combined gas law to relate pressure to
absolute temperature.
P1V1 P2 V2
=
T1
T2
(n constant)
P1 P2
=
T1 T2
At constant volume
Execute the plan:
(V and n constant)
T1 = 15 °C + 273 = 288 K
T2 = 35 °C + 273 = 308 K
T
(308 K)
= 4.00 atm
P2 = P1 × 2 = (3.74 atm) ×
T1
(288 K)
Check your answer: A gas with a higher temperature should exert a higher pressure.
38. Define the problem: Given the original volume, pressure, and temperature of a gas sample, the new
temperature, and the new pressure, determine the new volume of the sample.
Develop a plan: Convert the temperature to Kelvin. Use the combined gas law to relate pressure to
absolute temperature.
P1V1 P2 V2
=
T1
T2
Execute the plan:
(n constant)
T1 = 22 °C + 273 = 295 K
T2 = 42 °C + 273 = 315 K
T
P
(315 K) (165 mmHg)
×
= 501 mL
V2 = V1 × 2 × 1 = (754 mL) ×
T1 P2
(295 K) (265 mmHg)
Check your answer: The temperature fraction:
(315 K)
(295 K)
is larger than one, consistent with
(165 mmHg)
is
(265 mmHg)
smaller than one, consistent with decreasing the volume due to an increased pressure. Clearly these
two effects counteract each other, but this pressure change affects the volume more than the
temperature change does.
increasing the volume due to the increased temperature. The pressure fraction:
44. Define the problem: Given the mass of sucrose, the formula, and the product of a reaction, determine
the maximum volume of CO2 produced at STP. Compare that volume with the typical volume of two
loaves of French bread.
Develop a plan: Balance the equation. Determine the moles of sucrose from the molar mass, then use
the stoichiometric relationships given in the balanced equation to determine the moles of CO2
produced. Last, use the molar volume of a gas at STP to determine the volume of CO2. Estimate the
total volume of two loaves of French bread assuming they are cylinders. Compare the two volumes.
Execute the plan: Balance the equation:
C12H22O11(s) + 12 O2(g)
2.4 g C12H 22O11 ×
12 CO2 (g) + 11 H2O (l)
1 mol C12 H 22O11
12 mol CO2
22.414 L CO2 at STP
×
×
= 1.9 L
1 mol CO2
342.2956 g C12H 22O11 1 mol C 12H 22O11
CO2
Assume one loaf of French bread is a cylinder, 3.0 inches in diameter and 18 inches long.
18"
3"
r = 1.5 in
2
2
2
A = πr = π(1.5 in) = 7.1 in
2
V = Al = (7.1 in ) × (18 in) = 130 in
3
 2.54 cm 3 1 mL
1L
×
×
= 2.1 L
130 in  1 in 
1 cm 3 1000 mL
3 ×
Two loaves would have twice this volume: 2 × (2.1 L) = 4.2 L. The CO2 bubbles produced in the
bread are nearly half of its volume.
Check your answers: Slicing open French bread we see that it has a vast “honeycomb” of bubbleshaped spaces in it. It makes sense that approximately half the loaf’s volume can be associated with
the CO2 bubbles formed by the yeast when the bread was rising.
46. Define the problem: Given the balanced chemical equation for a chemical reaction and the volume of
one reactant at a specified pressure and temperature, determine the volume of the other reactant at a
specified pressure and temperature that will cause complete reaction, and the volume of one of the
products at a specified pressure and temperature that will be produced.
Develop a plan: Avogadro’s Law allows us to interpret a balanced equation with gas reactants and
products in terms of gas volumes, as long as their temperatures and pressures are the same. Use the
stoichiometry to relate liters of SiH4(g) that react with liters of O2(g) and liters of H2O(g).
Execute the plan: The balanced equation tells us that one volume of SiH4(g) reacts with two volumes
of O2(g) to make two volumes of H2O(g), since all the volumes are measured at the same temperature
and pressure.
2 LO2 (g)
= 10.4 LO2 (g)
5.2 L H 2 (g) ×
1 LSiH (g)
4
5.2 L H 2 (g) ×
2 L H 2O(g)
= 10.4 LH 2 O(g)
1 LSiH 4 (g)
Note: Multiplying a number by an exact whole number, n, is like adding that number to itself n times. 2
× (5.2) = 5.2 L + 5.2 L = 10.4 L, that is why we use the addition rule for assessing the significant
figures of the results here.
Check your answers: Twice as many O2 molecules are needed compared to the number of SiH4
molecules, forming twice as many H2O molecules. So, it makes sense that both the volume of O2 and
the volume of H2O are twice the volume of SiH4.
48. Define the problem: Given the balanced chemical equation for a chemical reaction, the mass of one
reactant, and excess other reactant, determine the pressure of the product produced at a specified
volume and temperature.
Develop a plan: Convert from grams to moles. Use the stoichiometric relationship from the balanced
equation to determine moles of product. Use the ideal gas law to determine the pressure of the
product.
Execute the plan:
0.050 g B 4 H10 ×
1 mol B 4 H 10
10 mol H 2 O
×
= 0.0047 mol H 2O
2
mol B 4 H10
53.32 g B4 H 10
T = 30. °C + 273 = 303 K

L⋅ atm 
(0.0047 mol H 2 O) × 0.08206
 × (303 K)
n H 2 O RT
mol ⋅ K 

=
= 0.027 atm
P=
V
(4.25 L)
0.027 atm ×
760 mmHg
= 21 mmHg
1 atm
Check your answer: The relative quantities of B4H10 and H2O seem sensible. All units cancel
properly in the calculation of atmosphere. This is a relatively low pressure for water but the sample is
also small.
52. (a) This equation is given in Problem-Solving Exercise 10.11:
2 C8H18(l) + 25 O2(g)
16 CO2(g) + 18 H2O(g)
(b) Define the problem: Given the length in miles of a trip, the fuel efficiency of a car, the density of
the liquid fuel, and the temperature and pressure, determine the volume of a gas–phase product
produced during the trip.
Develop a plan: Use the miles, the fuel efficiency, volume conversions, the density, and the
stoichiometry to find the moles of the product. Use the ideal gas law to determine the volume of
the product.
Execute the plan:
10. miles ×
1 gal gasoline 3.785 L gasoline 1000 mL 1 cm3
3
3
×
×
×
= 1.2 × 10 cm gasoline
32 miles
1 gal gasoline
1L
1 mL
1.2 ×10 3 cm3 gasoline ×
0.692 g gasoline
3
1 cm gasoline
×
1 mol gasoline
16 mol CO2
×
= 58 mol CO2
114.22 g gasoline 2 mol gasoline
T = 25 °C + 273 = 298 K

L ⋅atm 
(58 mol CO2 ) × 0.08206
 × (298 K)
nCO 2 RT
mol ⋅ K 

3
V=
=
= 1.4 ×10 L
P
(1.0 atm)
Check your answer: This is a large volume of CO2! However, it is not an unreasonable quantity
considering how many gallons of gasoline are used and how much CO2 is generated from each
octane molecule. Comparing the results to the answer in Question 53, less CO2 is generated by
methanol as a fuel than octane.
60. Define the problem: Given the partial pressure of several gases in a sample of the atmosphere, and the
total pressure of the sample, determine the partial pressure of O2, the mole fraction of each gas, and the
percent by volume. Compare the percentages to Table 10.3.
Develop a plan: Dalton’s law and its applications are described in Section 10.8. Dalton’s law of
partial pressures states that the total pressure (P) exerted by a mixture of gases is the sum of their
partial pressures (p1, p2, p3, etc.), if the volume (V) and temperature (T) are constant.
Ptot = p1 + p2 + p3 + ...
(V and T constant)
Because pi = XiPtot, we can use the total pressure and the partial pressure of a component to determine
its mole fraction:
Xi =
pi
Ptot
According to Avogadro’s law, moles and gas volumes are proportional, so the mole fraction is equal to
the volume fraction. To get percent, multiply the volume fraction by 100 %.
Execute the plan:
(a) Ptot = pN + pO + pAr + pCO + pH
2
2
2
2
pO = Ptot – pN – pAr – pCO – pH
2
2
2
2
O
O
= (740. mmHg) – (575 mmHg) – (6.9 mmHg) – (0.2 mmHg) – (4.0 mmHg) = 154 mmHg
(b) X N 2 =
pN
2 = 575 mmHg = 0.777
740. mmHg
Ptot
X O2 =
p
6.9 mmHg
X Ar = Ar =
= 0.0093
Ptot 740. mmHg
2 = 154 mmHg = 0.208
740. mmHg
Ptot
X CO 2 =
X H 2O =
(c)
pO
p CO
2 = 0.2 mmHg = 0.0003
740. mmHg
Ptot
pH O
2 = 4.0 mmHg = 0.0054
740. mmHg
Ptot
% N2 = XN × 100 % = 0.777 × 100 % = 77.7 %
2
% O2 = XO × 100 % = 0.208 × 100 % = 20.8 %
2
% Ar = XAr × 100 % = 0.0093 × 100 % = 0.93 %
% CO2 = XCO × 100 % = 0.0003 × 100 % = 0.03 %
2
% H2O = XH
2
O
× 100 % = 0.0003 × 100 % = 0.54 %
The Table 10.3 figures are slightly different. This sample is wet, whereas the proportions given in
Table 10.3 are for dry air.
Check your answers: The percentages are very close to those provided in the table, and the variations
are explainable. The sum of the mole fractions is 1, and the sum of the percentages is 100 %.
3
3
108. The initial volume is 40 cm and the final volume is 60 cm . This 2:3 ratio in the gas volumes
means for every two molecules of gas reactants there must be three molecules of gas products. The
reaction that exhibits this ratio is (c):
2 AB2(g)
A2(g) + 2 B2(g)
109. The initial volume is 1.8 L and the final volume is 0.9 L. This 2:1 ratio in the gas volumes means
for every two molecules of gas reactants there must be one molecule of gas products. The reactant
count is six, so the box that has three product molecules fits these observations. The correct box is
(b):
6 AB2(g)
3 A2B4(g)
116. Define the problem: Given the volume of liquid oxygen and its density, determine the volume of air
(a mixture of gases) at STP that is necessary produce this amount of liquid oxygen.
Develop a plan: Use the density and molar mass to determine the moles of the liquid. Then use the
molar volume of a gas at STP and the volume percent of oxygen in air to calculate the volume of dry
air.
Execute the plan:
200. L O2 ×
1000 mL 1.14 g O2
1 mol O 2
×
×
= 7.13 × 103 mol O2
1L
1 mL O2 31.9988 g O 2
The moles of O2 in the liquid must come from the air. Table 10.1 gives the percentage by volume
for oxygen in air.
3
7.13×10 mol O2 ×
22.414 L O2
100 Lair
5
×
= 7.62× 10 L air
1 mol O2 at STP 20.948 L O2
Check your answer: Gases have a density much smaller than liquids, so the volume of gas must be
much larger than the volume of liquid condensed from it.