Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
Mini Investigation: Avoiding a Messy Opening, page 575
Answers may vary. Sample answers:
A. The dissolved carbon dioxide came out of the solution resulting in an increase in gas pressure
around the sides and at the top and bottom of the can. We should be nervous about opening the
can because when we open it, the carbon dioxide gas will leave the can rapidly since it is at a
higher pressure than the surrounding air. This will result in a mess!
B. After shaking the pop in the bottle, I observed a lot of bubbles of gas forming. The bubbles of
gas formed around the sides and bottom of the bottle. The gas bubbles will result in a messy
situation because they will be forced out of the bottle by the unequal pressure, and they will also
push the pop out of the bottle, spraying it everywhere.
C. I proposed tapping the sides and bottom of the bottle to move most of the gas bubbles to the
top of the bottle.
D. After tapping the bottle and moving the gases toward the top, when the bottle was opened the
carbon dioxide gas escaped without affecting the pop underneath it. The action was successful
because the pop did not spray everywhere.
Section 12.1: Avogadro’s Law and Molar Volume
Tutorial 1 Practice, page 579
1.
Given: initial volume, V1 = 43 L
initial amount of gas, n1 = 2.4 mol
final amount of gas, n2 = 4.8 mol
The pressure and temperature remain constant.
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution: Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in
the known values, and solve the equation.
Vn
V2 = 1 2
n1
=
43 L ! 4.8 mol
2.4 mol
V2 = 86 L
Statement: The new amount of gas will occupy 86 L.
2. (a) Given: initial volume, V1 = 10.00 L
initial amount of gas, n1 = 1.80 mol
final amount of gas, n2 = n1 + nAr added
amount of argon added, nAr added = 1.80 mol
The pressure and temperature remain constant.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-1
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution:
Step 1. Determine the final total amount of argon, n2.
n2 = n1 + nAr added
= 1.80 mol + 1.80 mol
n2 = 3.60 mol
Step 2. Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in the
known values, and solve the equation.
Vn
V2 = 1 2
n1
=
10.00 L ! 3.60 mol
1.80 mol
V2 = 20.0 L
Statement: The new volume of the gas is 20.0 L.
(b) Given: initial volume, V1 = 10.00 L
initial amount of gas, n1 = 1.80 mol
final amount of gas, n2 = n1 + nAr added
mass of argon added, mAr added = 25.0 g
The pressure and temperature remain constant.
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution:
Step 1. Determine the molar mass of argon.
MAr = 39.95 g/mol
Step 2. Determine the amount of argon added, nAr added, using the appropriate conversion factor
derived from the molar mass of argon.
1 mol
nAr added = mAr added !
39.95 g
1 mol
= 25.0 g !
39.95 g
nAr added = 0.626 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-2
Step 3. Determine the final total amount of argon, n2.
n2 = n1 + nAr added
= 1.80 mol + 0.626 mol
n2 = 2.43 mol
Step 4. Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in the
known values, and solve the equation.
Vn
V2 = 1 2
n1
=
10.00 L ! 2.43 mol
1.80 mol
V2 = 13.5 L
Statement: The new volume of the gas is 13.5 L.
(c)
Given: initial volume, V1 = 10.00 L
initial amount of gas, n1 = 1.80 mol
final amount of gas, n2 = n1 – nAr escaped
half of the argon escaped, nAr escaped = half of n1
The pressure and temperature remain constant.
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution:
Step 1. Determine the final total amount of argon, n2.
1
nAr escaped = n1 !
2
1
= 1.80 mol !
2
nAr escaped = 0.900 mol
n2 = n1 ! nAr escaped
= 1.80 mol ! 0.900 mol
n2 = 0.900 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-3
Step 2. Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in the
known values, and solve the equation.
Vn
V2 = 1 2
n1
=
10.00 L ! 0.900 mol
1.80 mol
V2 = 5.00 L
Statement: The new volume of the gas is 5.00 L.
3. Given: initial volume, V1 = 12.0 L
initial mass of carbon dioxide, m1 = 4.80 g
molar mass of carbon dioxide, M Co = 44.0 g/mol
2
final amount of gas, n2 = n1 + nCo
2
added
amount of carbon dioxide added, nCo
2
added
= 0.5 mol
The pressure and temperature remain constant.
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution:
Step 1. Determine the initial amount of carbon dioxide present, n1, using the appropriate
conversion factor derived from the molar mass of argon.
1 mol
n1 = m1 !
44.0 g
1 mol
= 4.80 g !
44.0 g
nAr added = 0.109 mol
Step 2. Determine the final total amount of carbon dioxide, n2.
n2 = n1 + nAr added
= 0.109 mol + 0.5 mol
n2 = 0.61 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-4
Step 3. Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in the
known values, and solve the equation.
Vn
V2 = 1 2
n1
=
12.0 L ! 0.61 mol
0.109 mol
V2 = 7 ! 10 L
Statement: The new volume of the balloon is 7 × 10 L.
Tutorial 2 Practice, page 580
1. (a) Given: amount of nitrous oxide, n = 1.0 mol; STP conditions
Required: volume, V
22.4 L
Analysis: V = n !
1 mol
Solution: Substitute in the known values and solve the mathematical equation.
22.4 L
V = 1.0 mol !
1 mol
V = 22 L
Statement: The volume occupied by 1.0 mol of nitrous oxide gas is 22 L at STP.
(b) Given: amount of nitrous oxide, n = 2.0 mol; STP conditions
Required: volume, V
22.4 L
Analysis: V = n !
1 mol
Solution: Substitute in the known values and solve the mathematical equation.
22.4 L
V = 2.0 mol !
1 mol
V = 45 L
Statement: The volume occupied by 2.0 mol of nitrous oxide gas is 45 L at STP.
(c) Given: amount of nitrous oxide, n = 4.5 mol; STP conditions
Required: volume, V
22.4 L
Analysis: V = n !
1 mol
Solution: Substitute in the known values and solve the mathematical equation.
22.4 L
V = 4.5 mol !
1 mol
V = 101 L
Statement: The volume occupied by 4.5 mol of nitrous oxide gas is 101 L at STP.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-5
2. Given: volume, V = 145.6 L
pressure = 101.3 kPa
temperature = 0 °C
Required: amount of oxygen gas, n
Analysis: The pressure and temperature indicate that the conditions are STP, so the conversion
factor derived from the molar volume at STP may be used. In this case we wish to find the
amount of oxygen gas, n, so we use the factor 1 mol/22.4 L as follows:
1 mol
n=V !
22.4 L
Solution: Substitute in the known values and solve the mathematical equation.
1 mol
n = 145.6 L !
22.4 L
n = 6.500 mol
Statement: The amount of oxygen gas in the container is 6.500 mol.
3. Given: V = 44.8 L; STP conditions
Required: mass of hydrogen gas, mH
1 mol
Analysis: n = V !
22.4 L
2.02 g
MH =
2
1 mol
Solution:
Step 1. Determine the amount of hydrogen gas in the container by multiplying the volume of
hydrogen by an appropriate conversion factor derived from the molar volume at STP. In this
case, the necessary conversion factor is 1 mol/22.4 L.
1 mol
n=V !
22.4 L
1 mol
= 44.8 L !
22.4 L
n = 2.00 mol
Step 2. Determine the mass of hydrogen gas in the container by multiplying the amount of
hydrogen by an appropriate conversion factor derived from the molar mass of hydrogen.
2.02 g
mH = n !
1 mol
2.02 g
= 2.00 mol !
1 mol
mH = 4.04 g
Statement: The mass of hydrogen gas collected is 4.04 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-6
Section 12.1 Questions, page 581
1. Avogadro’s law states that the volume of a gas is directly proportional to the amount of gas
when the temperature and pressure of the gas remain constant. If the amount of gas in a balloon
V
is tripled, then the volume of the balloon must also triple in order for
to equal a constant
n
(assuming temperature and pressure are kept constant).
2.
Given: initial volume, V1 = 38.5 L
initial amount of gas, n1 = 2.5 mol
final amount of gas, n2 = 1.5 mol
The pressure and temperature remain constant.
Required: Identify the unknown variable.
V2 = ?
Analysis: Use Avogadro’s law to find the final volume of the gas.
V1 V2
=
n1 n2
Solution: Rearrange the Avogadro’s law equation to isolate the unknown variable, substitute in
the known values, and solve the equation.
Vn
V2 = 1 2
n1
=
38.5 L ! 1.5 mol
2.5 mol
V2 = 23 L
Statement: The volume of 1.5 mol of butane gas is 23 L.
3. (a) Given: volume, V = 44.8 L; SATP conditions
Required: amount of nitrogen gas, n
1 mol
Analysis: n = V !
24.8 L
Solution: Substitute in the known values and solve the mathematical equation.
1 mol
n = 44.8 L !
24.8 L
n = 1.81 mol
Statement: The amount of nitrogen in 44.8 L of pure gas is 1.81 L at SATP.
(b) Given: amount of propane gas, n = 4.8 mol; SATP conditions
Required: volume, V
24.8 L
Analysis: V = n !
1 mol
Solution: Substitute in the known values and solve the mathematical equation.
24.8 L
V = 4.8 mol !
1 mol
V = 120 L
Statement: The volume occupied by 4.8 mol of propane gas is 120 L at SATP.
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Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-7
(c) Given: V = 34.6 L; SATP conditions
Required: mass of carbon dioxide, mCO
2
Analysis: n = V !
1 mol
24.8 L
44.01 g
2
1 mol
Solution:
Step 1. Determine the amount of carbon dioxide gas in the sample by multiplying the volume of
carbon dioxide by an appropriate conversion factor derived from the molar volume at SATP. In
this case, the necessary conversion factor is 1 mol/24.8 L.
1 mol
n=V !
24.8 L
1 mol
= 34.6 L !
24.8 L
n = 1.3952 mol [two extra digits carried]
Step 2. Determine the mass of carbon dioxide gas in the container by multiplying the amount of
carbon dioxide by an appropriate conversion factor derived from the molar mass of carbon
dioxide.
M CO =
mCO = n !
2
44.01 g
1 mol
= 1.3952 mol !
44.01 g
1 mol
mCO = 61.4 g
2
Statement: The mass of carbon dioxide in 34.6 L of carbon dioxide gas at SATP is 61.4 g.
(d) Given: mCH = 1250 g ; SATP conditions
4
Required: volume of methane, V
24.8 L
Analysis: V = n !
1 mol
16.04 g
M CH =
4
1 mol
Solution:
Step 1. Determine the amount of methane gas in the sample by multiplying the mass of methane
by an appropriate conversion factor derived from the molar mass of methane.
1 mol
n = mCH !
4
16.04 g
= 1250 g !
1 mol
16.04 g
n = 77.9 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-8
Step 2. Determine the volume of methane gas in the sample by multiplying the amount of
methane by an appropriate conversion factor derived from the molar volume at SATP. In this
case, the necessary conversion factor is 24.8 L/1 mol.
24.8 L
V =n!
1 mol
24.8 L 103 mL
= 77.9 mol !
!
1 L
1 mol
103 mL
1 L
6
V = 1.93 ! 10 mL
Statement: The volume of 1250 g of methane is 1.93 × 106 mL at SATP.
(e) Given: V = 36.5 L; SATP conditions
Required: amount of oxygen gas, n
1 mol
Analysis: n = V !
24.8 L
16.00 g
MO =
2
1 mol
Solution:
Determine the amount of oxygen gas in the sample by multiplying the volume of oxygen by an
appropriate conversion factor derived from the molar volume at SATP. In this case, the
necessary conversion factor is 1 mol/24.8 L.
1 mol
n=V !
24.8 L
1 mol
= 36.5 L !
24.8 L
n = 1.47 mol
Statement: The amount of oxygen gas in 36.5 L of O2 gas at SATP is 1.47 mol.
4. (a) Given: mH = 4.50 g ; STP conditions
= 1930 L !
2
Required: volume hydrogen gas, V
22.4 L
Analysis: V = n !
1 mol
2.02 g
MH =
2
1 mol
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-9
Solution:
Step 1. Determine the amount of hydrogen gas collected by multiplying the mass of hydrogen by
an appropriate conversion factor derived from the molar mass of hydrogen.
1 mol
n = mH !
2
2.02 g
1 mol
= 4.50 g !
2.02 g
n = 2.228 mol [two extra digits carried]
Step 2. Determine the volume of methane gas in the sample by multiplying the amount of
methane by an appropriate conversion factor derived from the molar volume at STP. In this case,
the necessary conversion factor is 22.4 L/1 mol.
22.4 L
V =n!
1 mol
22.4 L
= 2.228 mol !
1 mol
V = 49.9 L
Statement: The volume of 4.50 g of hydrogen is 49.9 L at STP.
(b) Finding the amount of hydrogen gas (in moles) produced at STP from 0.52 mol of
magnesium completely reacting is a stoichiometry problem. There is a 1 : 1 mole ratio of
magnesium to hydrogen gas in the balanced chemical equation, so 0.52 mol of magnesium will
produce 0.52 mol of hydrogen gas at STP. The mass of 0.52 mol of hydrogen gas can be found
as follows:
Given: amount of hydrogen gas, n = 0.52 mol; STP conditions
Required: mass of hydrogen has, mH
2
Analysis: n = V !
1 mol
22.4 L
2.02 g
2
1 mol
Solution: Determine the mass of hydrogen gas produced by multiplying the amount of hydrogen
by an appropriate conversion factor derived from the molar mass of hydrogen.
2.02 g
mH = n !
2
1 mol
2.02 g
= 0.52 mol !
1 mol
mH = 1.1 g
MH =
2
Statement: The mass of 0.52 mol of hydrogen gas is 1.1 g.
Copyright © 2011 Nelson Education Ltd.
Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-10
5. Answers may vary. Sample answer: No, this is not a good example of Avogadro’s law. When
the basketball begins to fill, the situation is a close example of Avogadro’s law, since the
pressure and temperature are constant and the volume of the basketball increases directly
proportionally to the amount of air pumped in. But as the basketball fills, the pressure inside the
ball begins to change, and the volume does not increase at a rate that is directly proportional to
the amount of air pumped in. The volume begins increasing less and less as more gas is pumped
in and the air inside of the basketball is put under higher and higher pressure. This high pressure
is what allows the basketball to bounce.
6. Gases, such as barbecue propane gas, are sold by mass and not by volume because gases fill
the container they occupy. Different masses of the same gas can be pressurized so that they fill
the same volume, so selling gases by volume could result in customers receiving widely varying
masses of gas.
7. Answers may vary. Sample answer: John McLennan was born in 1867 near Woodstock,
Ontario. He studied at the University of Toronto, earning a Ph.D in physics in 1900. At the start
of World War I, John McLennan was commissioned by the British government to survey the
various countries under the British Empire for sources of helium gas. Since helium gas was rare,
it was very expensive. McLennan’s search took him to Bow Island, near Calgary, Alberta. He
discovered that Bow Island natural wells contained up to 0.36 % helium.
He set up facilities in Calgary for the extraction and purification of helium because
natural gas was plentiful. Helium extraction was achieved through a low-temperature
liquefaction process. He was able to do this on such a large scale that by 1919 the price of helium
had dropped significantly. At the end World War I, McLennan had collected 2000 m3 of helium
(approximately 90 % of which was pure). As the war ended, helium powered airships were being
replaced by gasoline powered airplanes. Rather than waste the pure helium, McLennan decided
to develop a cryogenic laboratory to produce liquid helium. He started a research program to
study this liquid at the University of Toronto.
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Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions
12.1-11