Section 12.5: Reactions of Gases and Gas Stoichiometry
Tutorial 1 Practice, page 599
1. Given: volume of hydrogen, VH = 54 L
2
Required: volume of oxygen, VO
2
Solution:
Step 1. Write the balanced chemical equation for the reaction, listing the given values and
required values.
2 H 2 (g) + O 2 (g) ! 2 H 2O(g)
54 L
VO
2
Step 2. Convert the volume of the given substance, VH , to the volume of the required substance,
2
VO .
2
VO = VH !
2
2
= 54 L !
1 molO
2
2 molH
2
1 molO
2
2 molH
2
VO = 27 L
2
Statement: The volume of oxygen required is 27 L.
2. Given: volume of ammonia, VNH = 34.5 L
3
Required: volume of hydrogen gas, VH
2
Solution:
Step 1. Write the balanced chemical equation for the reaction, listing the given values and
required values.
3 H 2 (g) + N 2 (g) ! 2 NH 3 (g)
VH
34.5 L
2
Step 2. Convert the volume of the given substance, VNH , to volume of the required substance,
3
VH .
2
VH = VNH !
2
3
3 molH
2
2 molNH
= 34.5 L !
3
3 molH
2
2 molNH
3
VH = 51.8 L
2
Statement: The volume of hydrogen gas required is 51.8 L.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-1
3. Given: volume of nitrogen, VN = 15.0 L
2
Required: volume of oxygen, VO
2
Solution:
Step 1. Write the balanced chemical equation for the reaction, listing the given values and
required values below.
4 KNO3 (s) ! 2 K 2O(g) + 2 N 2 (g) + 5 O 2 (g)
15.0 L
VO
2
Step 2. Convert the volume of the given substance, VN , to volume of the required substance,
2
VO .
2
5 molO
VO = VN !
2
2
2 molN
2
= 15.0 L !
2
5 molO
2
2 molN
2
VO = 37.5 L
2
Statement: The volume of oxygen required is 37.5 L.
Tutorial 2 Practice, page 602
1. Given: mass of oxygen gas, mO = 128 g
2
molar mass of oxygen gas, M O = 32.00 g/mol
2
The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa.
Required: volume of nitrogen dioxide, VNO
2
Solution:
Step 1. Write the balanced chemical equation, listing the given and required quantities.
2 NO(g) + O 2 (g) ! 2 NO 2 (g)
128 g
VNO
2
Step 2. Convert the mass of the given substance, mO , into an amount, nO .
2
nO = 128 g !
2
2
1 mol
32.00 g
nO = 4.00 mol
2
Step 3. Determine the amount of nitrogen dioxide produced from the amount of oxygen gas.
2 molNO
2
nNO = 4.00 mol !
2
1 molO
2
nNO = 8.00 mol
2
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-2
Step 4. Determine the volume of the required substance, VNO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol
PVNO = nNO RT
2
–1
–1
2
i K .
2
VNO =
2
=
nNO RT
2
P
(8.00 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 273 K
$
)
101.325 kPa
VNO = 179 L
2
Statement: When 128 g of oxygen gas reacts with nitrogen monoxide, 179 L of nitrogen dioxide
is produced (under the given conditions of temperature and pressure).
2. Given: volume of carbon dioxide, VCO = 2.00 L
2
temperature, T = 300.0 K
pressure, P = 101.3 kPa
Required: mass of sodium carbonate, mNa CO
2
3
Solution:
Step 1. Write the balanced chemical equation, listing the given and required quantities.
Na 2CO3 (s) + 2 HCl(aq) ! 2 NaCl(aq) + CO 2 (g) + H 2O(g)
mNa CO
2
2.00 L
3
Step 2. Determine the amount of the given substance, nCO , produced in the reaction using the
2
ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVCO = nCO RT
2
2
nCO =
2
=
PVCO
RT
2
(101.3 kPa )(2.00 L )
(8.314 kPa i L i mol
–1
)(
i K –1 300.0 K
)
nCO = 0.0812 mol
2
Step 3. Determine the amount of the sodium carbonate that is required to react with the amount
of carbon dioxide, nNa CO .
2
nNa CO = 0.0812 mol !
2
3
3
1 molNa CO
2
1 molCO
3
2
nNa CO = 0.0812 mol
2
3
Step 4. Determine the mass of sodium carbonate that is required from the amount of sodium
carbonate.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-3
M Na CO = 105.99 g/mol
2
3
105.99 g
1 mol
105.99 g
= 0.0812 mol !
1 mol
= 8.61 g
mNa CO = nNa CO !
2
3
mNa CO
2
3
2
3
Statement: 8.61 g of sodium carbonate is required to produce 2.00 L of carbon dioxide at these
conditions.
3. Given: volume of ethane, VC H = 82.0 L
2
6
temperature, t = 123 °C
pressure, P = 105 kPa
molar mass of ethane, M?
Required: volume of oxygen, VO
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 123 + 273
T = 396 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
2 C2 H 6 (g) + 7 O 2 (g) ! 6 H 2O (g) + 4 CO 2 (g)
82.0 L
VO
2
Step 3. Convert the volume of the given substance, VC H , into an amount, nC H , using the ideal
2
6
gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol
PVC H = nC H RT
2
6
2
nC H =
2
6
=
2
–1
6
i K .
6
PVC H
2
6
RT
(105 kPa )(82.0 L )
(8.314 kPa i L i mol
nC H = 2.6152 mol
2
–1
6
–1
)(
i K –1 396 K
)
[two extra digits carried]
Step 4. Determine the amount of oxygen required for the amount of ethane combusted.
7 molO
2
nO = 2.6152 mol !
2
2 molC H
2
nO = 9.1532 mol
2
6
[two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-4
Step 5. Determine the volume of the required substance, VO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol
PVO = nO RT
2
–1
2
–1
i K .
2
VO =
nNO RT
2
2
=
P
(9.1532 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 396 K
$
)
105 kPa
VO = 287 L
2
Statement: 287 L of oxygen is required for the combustion of 82.0 L of ethane at these
conditions.
Mini Investigation: Capturing Carbon Dioxide on a Small Scale, page 602
A. Answers may vary. Sample answer:
Ca OH 2 (aq) + CO 2 (g) ! CaCO3 (s) + H 2O
( )
B. Answers may vary. Sample answer: The formation of a white precipitate, which is the calcium
carbonate, supports the proposed chemical equation.
C. Bromothymol blue is blue in the basic solution formed by calcium oxide. Carbon dioxide is
an acidic oxide, so bubbling carbon dioxide into the mixture would form an acidic solution that
changes the colour of bromothymol blue from blue to yellow.
D. Answers may vary. Sample answer: Yes, this would be an effective method of removing
carbon dioxide from the atmosphere, as long as the amount of carbon dioxide released into the
atmosphere in the production of limewater is less than the amount captured by the process.
E. Answers may vary. Sample answer: An experiment could be conducted in which ambient air
is forced through the limewater and then the resulting solution is poured through a filter and the
calcium carbonate precipitate is collected. The steps could then be repeated except with forcing
exhaled air through the limewater. The two samples of precipitate collected could then be
massed and compared to determine how many times greater the concentration of carbon dioxide
in exhaled air is compared to ambient air.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-5
Section 12.5 Questions, page 603
1. (a) 2 C4 H10 (g) + 13 O 2 (g) ! 8 CO 2 + 10 H 2O
(b) Given: volume of butane, VC H = 5.60 L
4
10
The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa.
Required: volume of carbon dioxide, VCO
2
Solution:
Step 1. Write the balanced chemical equation, listing the given and required quantities.
2 C4 H10 (g) + 13 O 2 (g) ! 8 CO 2 + 10 H 2O
5.60 L
VCO
2
Step 2. Convert the volume of the given substance, VC H , into an amount, nC H .
4
10
4
10
1 mol
22.4 L
= 0.250 mol
nC H = 5.60 L !
4
10
nC H
4
10
Step 3. Determine the amount of carbon dioxide produced from the amount of butane gas used.
8 molCO
2
nCO = 0.250 mol !
2
2 molC H
4
10
nCO = 1.00 mol
2
Step 4. Determine the volume of the required substance, VCO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol
PVCO = nCO RT
2
–1
–1
2
i K .
2
VCO =
2
=
nCO RT
2
P
(1.00 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 273 K
$
)
101.325 kPa
VCO = 22.4 L
2
Statement: When 5.60 L of butane undergoes complete combustion at STP, 22.4 L of carbon
dioxide is produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-6
2. (a) Given: mass of potassium chlorate, mKClO = 122.6 g
3
molar mass of potassium chlorate, M KClO = 122.55 g/mol
3
The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa.
Required: volume of oxygen, VO
2
Solution:
Step 1. Write the balanced chemical equation, listing the given and required quantities.
2 KClO3 (s) ! 2 KCl(s) + 3 O 2 (g)
122.6 g
VO
2
Step 2. Convert the mass of the given substance, mKClO , into an amount, nKClO .
3
nKClO = 122.6 g !
3
3
1 mol
122.55 g
nKClO = 1.000 mol
3
Step 3. Determine the amount of oxygen gas produced from the amount of potassium chlorate
used.
3 molO
2
nO = 1.000 mol !
2
2 molKClO
3
nO = 1.500 mol
2
Step 4. Determine the volume of the required substance, VO , using the ideal gas law equation,
2
PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVO = nO RT
2
2
VO =
2
=
nO RT
2
P
(1.500 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 273 K
$
)
101.325 kPa
VO = 33.60 L
2
Statement: When 122.6 g of potassium chloride decomposes at STP, 33.60 L of oxygen gas is
produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-7
(b) Given: mass of potassium chlorate, mKClO = 122.6 g
3
molar mass of potassium chlorate, M KClO = 122.55 g/mol
3
The conditions are standard temperature and pressure, so t = 25 °C and P = 100 kPa.
Required: volume of oxygen, VO
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 25 + 273
T = 298 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
2 KClO3 (s) ! 2 KCl(s) + 3 O 2 (g)
122.6 g
VO
2
Step 3. Convert the mass of the given substance, mKClO , into an amount, nKClO .
3
nKClO = 122.6 g !
3
3
1 mol
122.55 g
nKClO = 1.000 mol
3
Step 4. Determine the amount of oxygen gas produced from the amount of potassium chlorate
used.
3 molO
2
nO = 1.000 mol !
2
2 molKClO
3
nO = 1.500 mol
2
Step 5. Determine the volume of the required substance, VO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol
PVO = nO RT
2
–1
2
–1
i K .
2
VO =
2
=
nO RT
2
P
(1.500 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 298 K
$
)
100 kPa
VO = 37.16 L
2
Statement: When 122.6 g of potassium chloride decomposes at SATP, 37.2 L of oxygen gas is
produced.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-8
3. Given: mass of sodium metal, mNa = 20.0 g
molar mass of sodium metal, M Na = 22.99 g/mol
temperature, t = 30 °C
pressure, P = 125 kPa
Required: volume of hydrogen gas, VH
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 30 + 273
T = 303 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
2 Na(s) + 2 H 2O(g) ! H 2 (g) + 2 NaOH
20.0 g
VH
2
Step 3. Convert the mass of the given substance, mNa , into an amount, nNa .
nNa = 20.0 g !
1 mol
22.99 g
nNa = 0.870 mol
Step 4. Determine the amount of hydrogen gas produced from the amount of sodium metal used.
1 molH
2
nH = 0.870 mol !
2
2 molNa
nH = 0.435 mol
2
Step 5. Determine the volume of the required substance, VH , using the ideal gas law equation,
2
PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVH = nH RT
2
2
VH =
2
=
nH RT
2
P
(0.435 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 303 K
$
)
125 kPa
VH = 9 L
2
Statement: When 20.0 g of sodium metal reacts with water, 9 L of hydrogen gas is produced
(under the given conditions of temperature and pressure).
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-9
4. (a) Given: mass of octane, mC H = 64.2 kg
8
18
molar mass of octane, M C H = 114.26 g/mol
8
18
temperature, t = 80 °C
pressure, P = 101.3 kPa
Required: volume of carbon dioxide, VCO
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 80 + 273
T = 353 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
2 C8 H18 (g) + 25 O 2 (g) ! 16 CO 2 (g) + 18 H 2O(g)
64.2 kg
VCO
2
Step 3. Convert the mass of the given substance, mC H , into an amount, nC H .
8
18
8
18
mC H = 64.2 kg
8
18
= 64.2 kg !
1000 g
1 kg
mC H = 6.42 ! 104 g
8
18
nC H = 6.42 ! 104 g !
8
18
nC H = 561.88 mol
8
18
1 mol
114.26 g
[two extra digits carried]
Step 4. Determine the amount of carbon dioxide produced from the amount of octane used.
16 molCO
2
nCO = 561.88 mol !
2
2 molC H
8
nCO = 4495.0 mol
2
18
[two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-10
Step 5. Determine the mass of carbon dioxide produced from the amount of carbon dioxide.
M CO = 44.01 g/mol
2
mCO = nCO !
2
2
44.01 g
1 kg
!
1 mol
103 g
= 4495.0 mol !
44.01 g
1 mol
1 kg
= 1.98 ! 105 g !
103 g
= 1.98 ! 102 kg
mCO = 2.0 ! 102 kg
2
Step 6. Determine the volume of the required substance, VCO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVCO = nCO RT
2
2
2
VCO =
2
=
nCO RT
2
P
(4495.0 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 353 K
$
)
101.3 kPa
VCO = 1.3 % 10 L
5
2
Statement: When a tank of 64.2 kg of gasoline undergoes combustion, 2.0 × 102 kg, or
1.3 × 105 L of carbon dioxide is produced (under the given conditions of temperature and
pressure).
(b) The mass of carbon dioxide produced from one tank of gas is 2.0 × 102 kg and each single
mature tree can absorb 22 kg of carbon dioxide per year, therefore
2.0 ! 102 kg
22 kg
=! 9 trees are
required to absorb the carbon dioxide produced from one tank of gas in one year.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-11
5. (a) Given: mass of calcium carbonate, mCaCO = 1025 g
3
molar mass of calcium carbonate, M CaCO = 100.09 g/mol
3
temperature, t = 20 °C
pressure, P = 100.0 kPa
Required: volume of carbon dioxide, VCO
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 20 + 273
T = 293 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
CaCO3 (s) + 2 HCl(aq) ! CaCl2 (aq) + H 2O(l) + CO 2 (g)
1025 g
VCO
2
Step 3. Convert the mass of the given substance, mCaCO , into an amount, nCaCO .
3
nCaCO = 1025 g !
3
3
1 mol
100.09 g
nCaCO = 10.24 mol
3
Step 4. Determine the amount of carbon dioxide produced from the amount of calcium carbonate
used.
1 molCO
2
nCO = 10.24 mol !
2
1 molCaCO
3
nCO = 10.24 mol
2
Step 5. Determine the volume of the required substance, VCO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVCO = nCO RT
2
2
2
VCO =
2
=
nCO RT
2
P
(10.24 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 293 K
$
)
100.0 kPa
VCO = 250 L
2
Statement: When 1025 g of calcium carbonate reacts with hydrochloric acid, 250 L of carbon
dioxide is produced (under the given conditions of temperature and pressure).
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-12
(b) Given: volume of carbon dioxide, VCO = 1550 L
2
temperature, T = 22.5 °C
pressure, P = 100.0 kPa
Required: mass of calcium carbonate, mCaCO
3
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 22.5 + 273
T = 295.5 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
CaCO3 (s) + 2 HCl(aq) ! CaCl2 (aq) + H 2O(l) + CO 2 (g)
mCaCO
1550 L
3
Step 3. Determine the amount of the given substance, nCO , produced in the reaction using the
2
ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVCO = nCO RT
2
2
nCO =
2
=
PVCO
RT
2
(100.0 kPa )(1550 L )
(8.314 kPa i L i mol
nCO = 63.091 mol
2
–1
)(
i K –1 295.5 K
)
[two extra digits carried]
Step 4. Determine the amount of the calcium carbonate that is required to react with the amount
of carbon dioxide.
1 molCaCO
3
nCaCO = 63.091 mol !
3
1 molCO
2
nCaCO = 63.091 mol
3
[two extra digits carried]
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-13
Step 5. Determine the mass of calcium carbonate that is required from the amount of calcium
carbonate.
M CaCO = 100.09 g/mol
3
mCaCO = nCaCO !
3
3
100.09 g
1 kg
!
1 mol
103 g
= 63.091 mol !
= 6.31 ! 103 g !
100.09 g
1 mol
!
1 kg
103 g
1 kg
103 g
mCaCO = 6.31 kg
3
Statement: 6.31 kg of calcium carbonate is required to produce 1550 L of carbon dioxide at
these conditions.
6. Given: mass of ammonium sulfate, m( NH ) SO = 264.0 g
4 2
4
molar mass of ammonium sulfate, M ( NH ) SO = 132.17 g/mol
4 2
4
mass of potassium hydroxide, mKOH = 280.0 g
molar mass of ammonium sulfate, M KOH = 56.11 g/mol
temperature, t = 23 °C
pressure, P = 64 kPa
Required: volume of ammonia gas, VNH
3
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 23 + 273
T = 296 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
NH 4 2 SO 4 (aq) + 2 KOH(aq) ! 2 NH 3 (g) + K 2SO 4 (aq) + 2 H 2O(l)
(
)
264.0 g
280.0 g
Copyright © 2011 Nelson Education Ltd.
VNH
3
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-14
Step 3. Convert the masses of the given substances, m( NH ) SO and mKOH , into amounts,
4 2
4
n( NH ) SO and nKOH .
4 2
4
1 mol
n( NH ) SO = 264.0 g !
4 2
4
132.17 g
n( NH ) SO = 1.997 mol
4 2
4
nKOH = 280.0 g !
1 mol
56.11 g
nKOH = 4.990 mol
Since the mole ratio of ammonium sulfate to potassium hydroxide is 1 : 2, and the amount of
ammonium sulfate is less than half of the amount of potassium hydroxide, the ammonium sulfate
will limit the reaction and it is this amount that will be used to determine the amount of ammonia
gas produced.
Step 4. Determine the amount of ammonia gas produced from the amount of ammonium sulfate
used.
2 molNH
3
nNH = 1.997 mol !
3
1 mol( NH ) SO
4 2
4
nNH = 3.994 mol
3
Step 5. Determine the volume of the required substance, VNH , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1.
PVNH = nNH RT
3
3
3
VNH =
3
=
nNH RT
3
P
(3.994 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 296 K
$
)
64 kPa
VNH = 150 L
3
Statement: When 264.0 g of ammonium sulfate reacts with 280.0 g of potassium hydroxide,
150 L of nitrogen dioxide is produced (under the given conditions of temperature and pressure).
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-15
7. Given: mass of propane, mC H = 54.0 g
3
8
molar mass of propane, M C H = 44.11 g/mol
3
8
temperature, t = 25 °C
pressure, P = 202.1 kPa
Required: volume of carbon dioxide, VCO
2
Solution:
Step 1. Convert the temperature value to kelvins.
T = t + 273
= 25 + 273
T = 298 K
Step 2. Write the balanced chemical equation, listing the given and required quantities.
C3H 8 (g) + 5 O 2 (g) ! 3 CO 2 (g) + 4 H 2O(g)
54.0 g
VCO
2
Step 3. Convert the mass of the given substance, mC H , into an amount, nC H .
3
nC H = 54.0 g !
3
8
8
3
8
1 mol
44.01 g
nC H = 1.2270 mol
3
8
[two extra digits carried]
Step 4. Determine the amount of carbon dioxide produced from the amount of propane used.
3 molCO
2
nCO = 1.2270 mol !
2
1 molC H
3
nCO = 3.6810 mol
2
8
[two extra digits carried]
Step 5. Determine the volume of the required substance, VCO , using the ideal gas law equation,
PV = nRT. Remember that R = 8.314 kPa i L i mol
PVCO = nCO RT
2
–1
–1
2
i K .
2
VCO =
2
=
nCO RT
2
P
(3.6810 mol ) !" 8.314 kPa i L i mol
–1
(
i K –1 # 298 K
$
)
202.1 kPa
VCO = 45.1 L
2
Statement: When 54.0 g of propane reacts with excess oxygen, 45.1 L of carbon dioxide is
produced (under the given conditions of temperature and pressure).
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.5-16