CHAPTER 10 GAS MIXTURES AND REACTIONS - Quia

CHAPTER 10 GAS MIXTURES AND REACTIONS

Try This Activity: Producing a “Natural” Gas

(Page 459)

• White cloudy mixture formed with many gas bubbles rising to the surface. Many droplets of liquid were spitting upfrom the surface.

• The lit match was extinguished.

• When the glass was tipped to “pour” the gas over the lit match, the match was also extinguished.

(a) Nitrogen, oxygen, and small quantities of other gases are present in the air.(b) Carbon dioxide is likely present after the reaction, because the gas does not support combustion (a characteristic of

carbon dioxide).(c) If the gas is bubbled through limewater, and the limewater turns cloudy, then carbon dioxide is likely present.(d) It suggests that the gas is more dense than air.(e) Carbon dioxide is used as a fire extinguisher (and also as an effervescent).(f) Carbon dioxide escapes from underground pockets of gas. Natural decomposition of limestone produces bubbles of

carbon dioxide that escape from the surface of lakes. (Other examples include volcanoes and some hot springs.)

10.1 MIXTURES OF GASES

PRACTICE

(Page 461)

Understanding Concepts1. The pressure of CO2(g) is constant at 2 kPa.2. (a) The total pressure is (593.3 + 157 + 11 + 0.5) mm Hg = 762 mm Hg, or the total pressure is (79.11 + 20.9 + 1.5

+ 0.07) kPa = 101.6 kPa.

(b) 762 mm� Hg� � �170610.3m2m�5 k

HPg�a

� � 102 kPa

The pressures of 760 mm Hg and 102 kPa (which is 101.6 kPa rounded to 3 significant digits) are equivalent.3. The partial pressure of helium is (14.0 – 1.1) atm = 12.9 atm.

Reflecting4. (a) Both alloys and the atmosphere are solutions — homogeneous mixtures.

(b) Alloy components are measured by mass, and the law of conservation of mass serves the same purpose for alloysthat Dalton’s law does for gas solutions.

PRACTICE

(Page 463)

Understanding Concepts5. (a) Pressure exerted by a gas is the total force per unit area of the gas molecules striking the interior surface of the

container.(b) Particles move in straight lines unless deviated by some outside force. Intermolecular forces are very small (negli-

gible) for most gases, so the molecules travel in (essentially) straight lines.6. (a) The pressure is directly proportional to the number of gas molecules present — so it should double.

(b) The pressure should still double — the kind of gas molecule is unimportant.(c) Dalton’s law of partial pressures is based on the concept that the kind of molecule is unimportant — the pressure

depends only on the relative numbers of molecules. 7. If gases react, the number and kind of molecules, and therefore the pressure, will change because of the reaction.

Copyright © 2002 Nelson Thomson Learning Chapter 10 Gas Mixtures and Reactions 287

288 Unit 4 Gases and Atmospheric Chemistry Copyright © 2002 Nelson Thomson Learning

PRACTICE

(Page 465)

Understanding Concepts8. From Table 3, the vapour pressure of water at 23°C is 2.81 kPa, so that will be the partial pressure inside the container.9. Use the Table 3 value for water vapour pressure at 20°C of 2.34 kPa. The partial pressure of nitrogen will be (98.1 –

2.34) kPa = 95.8 kPa.10. (a) Calculate as before, using Table 3. The partial pressure of hydrogen will be (92.4 – 3.17) kPa = 89.2 kPa.

(b) p1 = 89.2 kPa

v1 = 275 mL

p2 = 100 kPa

v2 = ?

p1v1 = p2v2

v2 = �p

p1v

2

1�

=

v2 = 245 mL

or

v2 = 275 mL � �8190.02

kkPPa�a�

�

v2 = 245 mL

The final volume of hydrogen would be 245 mL.

Applying Inquiry Skills

11. Experimental DesignAmmonia is collected by downward displacement of air in a fume hood, since it is much less dense than air.

Note: In practice, ammonia liquifies easily under only moderate pressure because of hydrogen bonding — soit is usually collected and transported as a liquid, called anhydrous (waterless) ammonia.

Making Connections12. The principal gases above the liquid are carbon dioxide and water vapour. The total pressure is noticeably higher than

atmospheric, so must be significantly greater than 100 kPa.

SECTION 10.1 QUESTIONS

(Page 465)

Understanding Concepts1. (a) The partial pressure of O2(g) is (385 – 240) kPa = 145 kPa.

(b) We assume that the total pressure is the sum of the partial pressures.2. (a) The total pressure in is (79.3 + 21.3 + 0.040 + 0.67) kPa = 101.3 kPa, and the total pressure out is (75.9 + 15.5

+ 3.7 + 6.2) kPa = 101.3 kPa.

(b) 101.3 kPa� � �101

1.3

a2t5m

kPa�� = 1.000 atm

(c) Since your lungs are open to the atmosphere when breathing, inhaled and exhaled air must begin and end at thesame (ambient atmospheric) pressure. The change in proportions of the gases are biological evidence that animalsneed oxygen to live — and chemical evidence that reactions have occurred to produce carbon dioxide and watervapour.

89.2 kPa� � 275 mL��

3. Dalton’s law works well for any gases that behave similarly to the “ideal” gas — that have small molecules and lowintermolecular forces, and do not react.

4. Dalton’s law can be explained by two concepts: gas particles act independently; and pressure is caused by particlecollisions with the walls of the container.

5. Dalton’s law works perfectly only for “ideal” gases; but to three significant digits, for most common gases, it workswell.

Making Connections6. Possibilities include the gradual absorption of emitted gases by physical, geological, and biological processes. For

example, carbon dioxide dissolves in water (oceans), and is used by plants to generate sugars through photosynthesis.The carbon in the gas can be followed through biological processes until it is deposited as sediment and buried, even-tually forming limestone.

Reflecting7. Diagrams drawn by students should show the principle illustrated in Figure 2, in a similar fashion. The adding of

amounts must be shown as proportional to the adding of pressures.Visual models are much easier for most people to understand than mathematical models.

10.2 REACTIONS OF GASES

PRACTICE

(Page 468)

Understanding Concepts1. Avogadro’s theory was needed to relate reacting volumes to equations.2. Avogadro used empirical observations of coefficient values from equations, and reacting volume ratios of gases.3. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

5.00 L v

Pressure and temperature conditions equal for all gases measured.

vO2= 5.00 L � �

51

�

vO2= 25.0 L

or Note: The text example on page 468 uses a mole ratio for conversion, but the volume ratio shown here is moreappropriate.

vO2= 5.00 L� C3H8� � �

1

5

L�L

C

O

3H2

8��

vO2= 25.0 L

The volume of oxygen required is 25.0 L.4. 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)

1.2 L v


vN2= 1.2 L � �

12

�

vN2= 0.60 L

or

vN2= 1.2 L� NO� � �

2

1

L�L

N

N

O�2�

vN2= 0.60 L

The volume of nitrogen produced is 0.60 L.


5. (a) 16 H2S(g) + 8 SO2(g) → 3 S8(s) + 16 H2O(g)

248 kL vAll gases measured at 250 kPa and 350°C.

vSO2= 248 kL � �

186�

vSO2= 124 kL

or

vSO2= 248 kL� H2S� � �

1

8

6

L

L�S

H

O

2

2

S��

vSO2= 124 kL

The volume of sulfur dioxide needed is 124 kL.(b) S8(s) + 8 O2(g) → 8 SO2(g)

v 250 kL

All gases measured at 200 kPa and 450°C.

vO2= 250 kL � �

88

�

vO2= 250 kL

or

vO2= 250 kL� SO2� � �

8

8

L�L

S

O

O2

2��

vO2= 250 kL

The volume of oxygen required is 250 kL.(c) 2 SO2(g) + O2(g) → 2 SO3(g)

v v 325 kL

All gases measured at the same temperature and pressure conditions.

vSO2= 325 kL � �

2

2�

vSO2= 325 kL

or

vSO2= 325 kL� SO3� � �

2

2

L�L S

S

O

O2

3��

vSO2= 325 kL

The volume of sulfur dioxide required is 325 kL.

vO2= 325 kL � �

12

�

vO2= 163 kL

or

vO2= 325 kL� SO3� � �

2

1

L�L

S

O

O2

3��

vO2= 163 kL

The volume of oxygen required is 163 kL.

Reflecting6. The law of combining volumes is similar to the law of definite proportions in that it shows combination in integer

ratios. It differs in that mass, and not volume, is compared; and also in that it does not apply to different substancesin reactions, but only to compound substances formed from the same elements.


PRACTICE

(Page 471)

Understanding Concepts7. vSO2

= 50 mL

VSATP = 24.8 L/mol

nSO2= ?

nSO2= �

50 m2L�4.

�

8 L�1 mol

�

nSO2= 2.0 mmol

The amount of sulfur dioxide is 2.0 mmol.

8. (a) nNe = 2.25 mol

VSTP = 22.4 L/mol

vNe = ?

vNe = 2.25 mol�� 212m.4

oLl�

�

vNe = 50.4 L

The volume of neon at STP is 50.4 L.(b) p1 = 101.325 kPa

T1 = 0°C = 273 K

p2 = ?

T2 = 35°C = 308 K

�T

p1

1� = �

T

p2

2�

p2 = �p

T1T

1

2�

=

pNe = 114 kPa

or

pNe = 101.325 kPa � �320783

K�K�

�

pNe = 114 kPa

The final pressure of the 50.4 L of warmed neon gas will be 114 kPa.(c) The tube must be transparent to allow light from the neon to escape, and strong enough to hold the pressure of the gas

when heated.9. mCO2

= 0.13 g

VSATP = 24.8 L/mol

MCO2= 44.01 g/mol

vCO2= ?

nCO2= 0.13 g� � �

414.

m01

olg�

�

nCO2= 0.0030 mol

101.325 kPa � 308 K��


vCO2= 0.0030 mol��

214m.8

oLl�

�

vCO2= 0.073 L = 73 mL

or

vCO2= 0.13 g� � �

414.

m01

ol�g�

� � �214m.8

oLl�

�

vCO2= 0.073 L = 73 mL

The volume of carbon dioxide at SATP is 73 mL.10. mC8H18

= 50.0 g

VSTP = 22.4 L/mol

MC8H18= 114.26 g/mol

vC8H18= ?

nC8H18= 50.0 g� � �

1114m.2

o6lg�

�

nC8H18= 0.438 mol

vC8H18= 0.438 mol� � �

212m.4

oLl�

�

vC8H18= 9.80 L

or

vC8H18= 50.0 g� � �

1114m.2

o6l�g�

� � �212m.4

oLl�

�

vC8H18= 9.80 L

The volume of octane vapour at STP would be 9.80 L.

11. m NO2= 1.00 Mg

VSATP = 24.8 L/mol

M NO2= 46.01 g/mol

v NO2= ?

n NO2= 1.00 Mg� � �

416.

m01

olg�

�

n NO2= 0.0217 Mmol = 21.7 kmol

v NO2= 21.7 kmol� � �

214m.8

oLl�

�

vNO2= 539 kL

orv NO2

= 1.00 Mg� � �416.

m01

ol�g�

� � �214m.8

oLl�

�

v NO2= 0.539 ML = 539 kL

The volume of nitrogen dioxide at SATP would be 539 kL or 539 m3.

12. mH2O = ?

v = 1.00 L

VSATP = 24.8 L/mol


M H2O = 18.02 g/mol

n H2O = 1.00 L� � �214m.8

oL�l

�

n H2O = 0.0403 mol

m H2O = 0.0403 mol�� 118.

m02

ol�g

�

m H2O = 0.727 g

or

m H2O = 1.00 L� � �214m.8

oL�l�

� � �118.

m02

ol�g

�

m H2O = 0.727 g

The mass of water required would be 0.727 g.

Applying Inquiry Skills13. (a) Prediction

The molar volume of oxygen at STP will be 22.4 L/mol, just as for any other gas, according to Avogadro’s theory.(b) Analysis

mO2= (46.84 – 45.79) g = 1.05 g

p1 = (95.2 – 2.64) kPa = 92.6 kPa

v1 = 848 mL

T1 = 22°C = 295 K

p2 = 101.325 kPa

vO2= ?

T2 = 0°C = 273 K

MO2= 32.00 g/mol

VSTP = ?

nO2= 1.05 g� � �

321.0m0ogl�

�

nO2= 0.0328 mol = 32.8 mmol

�p

T1v

1

1� = �p

T2v

2

2�

vo2

= �T

T2

1

p

p1

2

v1�

=

vo2

= 717 mL

or

vo2

= 848 mL � �10

912.3.625

kPka�Pa�

� � �227935

K�K�

�

vo2

= 717 mL

273 K� � 92.6 kPa� � 848 mL��

295 K� � 101.325 kPa�


The volume of oxygen at STP would be 717 mL. Since this is an amount of 32.8 mmol, the molar volume can now becalculated as follows:

VSTP = �32

7.187m�m�mLol

�

VSTP = 21.9 L/mol

According to the evidence from this investigation, the molar volume of oxygen at STP is 21.9 L/mol.(c) Evaluation

The prediction is judged to be verified, because the value calculated from the evidence is in close agreement.

difference = 22.4 – 21.9 L/mol = 0.5 L/mol

% difference = �202.5.4

LL//mm

ool�l�

� � 100% = 2%

(d) EvaluationAvogadro’s theory is supported by the result of this investigation, because the result agrees well with the predic-tion made from this authority.

Making Connections14. (a) MCO2

= 44.10 g/mol

VSATP = 24.8 L/mol

densityCO2= ?

densityCO2= �

4

2

4

4

.

.

1

8

0

L

g

/

/

m

m

o

o

l�l�

�

densityCO2= 1.8 g/L (to 2 significant digits)

The density of carbon dioxide at SATP is 1.8 g/L.

(b) Carbon dioxide is denser than air, and so will stay near the base of a fire to displace oxygen to extinguish the fire.(c) Carbon dioxide also works as an extinguisher because it will not support combustion.


(Page 474)

Understanding Concepts1. (a) Avogadro’s theory states that equal volumes of gases at equal temperature and pressure have equal numbers of

particles.(b) Avogadro’s theory provides the concept of a molar volume for substances in the gas phase, allowing chemical

calculations to be made easily for reactions with components that are gaseous.2. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

100 L v v v

All gases measured at 800°C and 200 kPa.

vO2= 100 L � �

54

�

vO2= 125 L

or

vO2= 100 L� NH3� � �

4

5

L�L

N

O

H2

3��

vO2= 125 L

The volume of oxygen required is 125 L.


vNO = 100 L � �44

�

vNO = 100 L

or

vNO = 100 L� NH3� � �44

L�L

NN

HO

3��

vNO = 100 L

The volume of nitrogen oxide produced is 100 L.

vH2O = 100 L � �64

�

vH2O = 150 L

or

vH2O = 100 L� NH3� � �4

6

L�L H

N2

H

O

3��

vH2O = 150 L

The volume of water vapour produced is 150 L.

(b) 2 NO(g) + O2(g) → 2 NO2(g)

v 750 L

All gases measured at 800°C and 200 kPa.

vO2= 750 L � �

12

�

vO2= 375 L

or

vO2= 750 L� NO2� � �

2

1

L�L

N

O

O2

2��

vO2= 375 L

The volume of oxygen required is 375 L.(c) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

100 L v

All gases measured at the same temperature and pressure.

vNO= 100 L � �13

�

vNO= 33.3 L

orvNO= 100 L� NO2� � �

31

L�L

NN

OO

2��

vNO= 33.3 L

The volume of nitrogen monoxide produced is 33.3 L.(d) No prediction can be made from the law of combining volumes, because the nitric acid and the ammonium nitrateare in aqueous solution, not in gaseous form.

3. (a) nH2= 7.50 mol

vH2= ?

VSATP = 24.8 L/mol

vH2= 7.50 mol��

214m.8

oLl�

�

vH2= 186 L

The volume occupied by the hydrogen is 186 L.


(b) p1 = 100 kPa

T1 = 25ºC = 298 K

v1 = 186 L

p2 = 1.2 kPa

T2 = –47ºC = 226 K

vH2= ?

�p

T1v

1

1� = �p

T2v

2

2�

vH2= �

T

T2p

1p1v

2

1�

=

vH2= 1.8 × 104 L = 18 kL

or

vH2= 186 L � �

110.20

kkPPa�a�

� � �222968

K�K�

�

vH2= 1.8 × 104 L = 18 kL

The final volume of hydrogen in the balloon will be 18 kL.

4. vO2= 20% of 20.0 L = 4.0 L

VSTP = 22.4 L/mol

nO2= ?

nO2= 4.0 L� � �

212m.4

oL�l

�

nO2= 0.18 mol

The amount of oxygen present is 0.18 mol.5. m CO2

= 1.00 t = 1.00 Mg

VSTP = 22.4 L/mol

M CO2= 44.01 g/mol

v CO2= ?

n CO2= 1.00 Mg� � �

414.

m01

olg�

�

n CO2= 0.0227 Mmol = 22.7 kmol

v CO2= 22.7 kmol��

212m.4

oLl�

�

v CO2= 509 kL

or

v CO2= 1.00 Mg� � �

414.

m01

ol�g�

� � �212m.4

oLl�

�

v CO2= 0.509 ML = 509 kL

The volume of carbon dioxide at STP would be 509 kL, or 509 m3.6. mO2

= ?

vO2= 1.9 kL

226 K� � 100 kPa� � 186 L��

298 K� � 1.2 kPa�


VSATP = 24.8 L/mol

MO2= 32.00 g/mol

nO2= 1.9 kL� � �

214m.8

oL�l

�

nO2= 0.077 kmol

mO2= 0.077 kmol��

3

1

2.

m

00

ol�g

�

mO2= 2.5 kg

ormO2

= 1.9 kL� � �214m.8

oL�l�

� � �3

1

2.

m

00

ol�g

�

mO2= 2.5 kg

The mass of oxygen consumed would be 2.5 kg.

10.3 THE OZONE LAYER

PRACTICE

(Page 479)

Understanding Concepts1. Ozone intercepts mostly the highest-energy (shorter wavelength) UV radiation from the Sun. Some UV radiation is

absorbed by oxygen to become ozone, and some UV radiation is absorbed by ozone decomposing.2. CFCs were developed as stable, non-toxic refrigerants, aerosol propellants, and foaming agents.3. In the upper stratosphere, CFCs initiate reactions that increase the rate of decomposition of ozone.4. An ozone “hole” is a (misleading) name for a region of very low ozone concentration; it is not a region where there

is no ozone.5. uv

CF3 + Br(g) → CFBr2(g) + Br(g)

Br(g) + O3(g) → BrO(g) + O2(g)

BrO(g) + O(g) → Br(g) + O2(g)

6. Ozone depletion is less severe in the Arctic because it is not as cold as the Antarctic, and because there is more airmixing due to prevailing winds.

7. Suntanning time should be decreased proportionally to a drop in the level of ozone in the stratosphere. Currentmedical thinking is that there is no absolutely “safe” level of sunlight exposure; so the perceived benefits of outdooractivities — and particularly of deliberate tanning — must be weighed against the increased risks of skin damage andskin cancer.


(Page 480)

Understanding Concepts1. The Montreal Protocol is an agreement among nations to decrease the production and use of CFCs, to try to prevent

damage to stratospheric ozone.2. Freeon-12 is a CFC refrigerant that is routinely recycled.3. HFE can replace CFCs for many uses, and hydrocarbons can be used as refrigerants.4. Canada’s Arctic Observatory and National Research Council contribute to research on effects of CFCs and the

development of alternative substances.5. Stratospheric ozone helps us by filtering potentially harmful UV radiation; but at low levels of the atmosphere (in the

air we breathe) ozone is dangerous — a very reactive and toxic substance.


Jim

Rectangle

10.4 GAS STOICHIOMETRY

PRACTICE

(Page 483)

Understanding Concepts1. 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)

15 g v

32.05 g/mol 22.4 L/mol (STP)

nCH3OH = 15 g� � �312.

m05

olg�

�

nCH3OH = 0.47 mol

nO2= 0.47 mol � �

32

�

nO2= 0.70 mol

vO2= 0.70 mol��

212m.4

oLl�

�

vO2= 16 L

or

vO2= 15 g� CH3OH� � �

3

1

2

m

.05

ol�g�C

C

H

H3

3

O

O

H�H�

� � �2

3

m

m

ol�o

C

l�H

O

3

2�OH�

� � �2

1

2

m

.4

o

L

l� O

O

2�2�

vO2= 16 L

The volume of oxygen needed is 16 L2. 2 NaCl(l → 2 Na(s) + Cl2(g)

105 kg v

22.99 g/mol 24.8 L/mol (SATP)

nNa = 105 kg� � �212.

m99

olg�

�

nNa = 4.57 kmol

nCl2= 4.57 kmol � �

12

�

nCl2= 2.28 kmol

vCl2= 2.28 kmol��

214m.8

oLl�

�

vCl2= 56.6 kL

or

vCl2= 105 kgg� Na� � �

212.

m99

ol�g�NNa�a�

� � �1

2

m

m

o

o

l�l�C

N

l

a�2�

� � �2

1

4

m

.8

o

L

l� C

C

l

l

2�2�

vCl2= 56.6 kL

The volume of chlorine produced is 56.6 kL, or 56.6 m3.3. C3H8(g) + 5 O2(g) → 3 CO2(g) + 8 H2O(g)

m v

44.11 g/mol 24.8 L/mol (SATP)

vO2= 125 L � 20% � 25 L

nO2= 25 L� � �

214m.8

oL�l

�


nO2= 1.0 mol

nC3H8= 1.0 mol � �

15

�

nC3H8= 0.20 mol

mC3H8= 0.20 mol��

414.

m11

ol�g

�

mC3H8= 8.9 g

or

mC3H8= 25 L� O2� � �

2

1

4

m

.8

o

L�l� O

O2�

2��

1

5

m

m

o

o

l�l�C

O3H

2�8�

� � �4

1

4.

m

11

ol�g

C

C

3

3

H

H

8�8�

mC3H8= 8.9 g

The mass of propane that can be burned is 8.9 g.


2 H2O2(aq) → 2 H2O(l) � O2(g)

50.0 mL v (lab p & T conditions)

0.88 mol/L

nH2O2= 50.0 mL� � �

0.818

L�mol�

nH2O2= 44 mmol

nO2= 44 mmol � �

12

�

nO2= 22 mmol

p = (94.6 – 2.49) kPa = 92.1 kPa (corrected for water vapour pressure)

T = 21°C = 294 K

vO2= ?

R = 8.31 kPa � L/(mol � K)

pv = nRT

vO2= �

nRp

T�

=

vO2= 5.8 �103 mL = 0.58 L

or

vO2= 50.0 mL� H2O2� ��

0.8

1

8

L�m

H

ol�

2O

H

2�2O2�

�� 2

1

m

m

o

o

l�l�H

O

2O2�

2��

8.

1

31

m

k

o

P

l�a�O

�

2�L

� K�O2��

922.914

kK�Pa�

�

vO2= 5.8 �103 mL = 0.58 L

According to the ideal gas law, the volume of oxygen at room pressure and temperature is predicted to be 0.58 L.

(b) AnalysisAccording to the evidence, the volume of oxygen gas produced at room conditions is 556 mL = 0.556 L.

(c) Evaluation

difference = 0.556 – 0.58L = 0.02 L

22 mmol� � �8.3

m1okl�P�

a�K�� L

� � 294 K��

92.1 kPa�


% difference = �00..0528

L�L�

� � 100% = 4%

The evidence agrees well with the predicted value, within 4%.

(d) EvaluationThe ideal gas law is judged to be verified by this investigation, since the evidence agrees well with the predic-

tion.

Making Connections5. (a) 2 H2(g) � O2(g) → 2 H2O(g)

300 L

All gases measured at 40°C and 1.50 atm.

vO2= 300 L � �

12

�

vO2= 150 L

or

vO2= 300 L� H2� � �

2

1

L�L O

H2

2��

vO2= 150 L

The volume of oxygen required is 150 L.(b) A vehicle burning hydrogen fuel by using oxygen from the air (which contains nitrogen) could still produce NOx

pollutants if the combustion temperature were high enough.


(Page 486)

Understanding Concepts1. 2 Fe(s) � 3 H2SO4(aq) → Fe2(SO4)3(aq) � 3 H2(g)

10 g v

55.85 g/mol 22.4 L/mol (STP)

nFe = 10 g� � �515.

m85

olg�

�

nFe = 0.18 mol

nH2= 0.18 mol � �

32

�

nH2= 0.27 mol

nH2= 0.27 mol��

212m.4

oLl�

�

nH2= 6.0 L

or

nH2= 10 g� Fe� � �

515.

m85

ol�g�FFe�e�

� � �3

2

m

m

o

o

l�l�

H

Fe�2��

2

1

2

m

.4

o

L

l� H

H

2�2�

nH2= 6.0 L

The volume of hydrogen produced will be 6.0 L.


2. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

2.00 ML, 0°C, 120 kPa v (SATP)

vO2= 2.00 ML � �

21

�

vO2= 4.00 ML (at 0°C = 273 K, 120 kPa)

p1 = 120 kPa

T1 = 0ºC = 273 K

v1 = 2.00 ML

p2 = 100 kPa

T2 = 25ºC = 298 K

vO2= ?

�p

T1v

1

1� = �p

T2v

2

2�

vO2= �

T

T2p

1p1v

2

1�

=

vO2= 2.62 ML

or

vO2= 2.00 ML� CH4��

1

2

L�L

C

O

H2

4��

112000

kkPP

a�a�

� � �229783

K�K�

�

vO2= 2.62 ML

The SATP volume of oxygen required is 2.62 ML, or 2.62 � 103 m3.

3. 2 NH3(g) � H2SO4(aq) → (NH4)2SO4(s)

75.0 kL, 10°C = 283 K, 110 kPa m

R = 8.31 kPa � L/(mol � K) 132.16 g/mol

pv = nRT

nNH3= �

Rpv

T�

=

nNH3= 3.51 kmol

n(NH4)2SO4= 3.51 kmol � �

12

�

n(NH4)2SO4= 1.75 kmol

m(NH4)2SO4= 1.75 kmol� � �

1312m.1

o6l�g

�

m(NH4)2SO4= 232 kg

or

m(NH4)2SO4= 75.0 kL� NH3� � � �

120803kK�Pa�

�1 mol� NH3��

��8.31 kPa� � 1 L� NH3�

110 kPa� � 75.0 kL��8.

131

mkoPla�� K�

� L�� 283 K�

298 K�� 120 kPa� � 2.00 ML��

273 K� � 100 kPa�


(continued) � �

m(NH4)2SO4= 232 kg

The mass of ammonium sulfate that can be produced is 232 kg.4. CH4�6H2O(s) → CH4(g) � 6 H2O(l)

1.0 kg v (20°C = 293 K, 95 kPa)

124.17 g/mol

R = 8.31 kPa � L/(mol � K)

nCH4�6H2O = 1.0 kg� � �12

14m.1

o7lg�

�

nCH4�6H2O = 0.00805 kmol

nCH4= 0.00805 kmol � �

11

�

nCH4= 0.00805 kmol

pv = nRT

vCH4= �

nRp

T�

=

vCH4= 0.21 kL

or

vCH4= 1.0 kg� CH4�6H2O� � � �

1 m

1

ol�m

C

o

H

l� C

4�

H

6H�4�

2O�

(continued) � �8.

1

31

m

k

o

P

l�a�C

�

H

L

4� �

C

K�H4� � �

92593

kPK�a�

�

vCH4= 0.21 kL

The volume of methane gas produced is 0.21 kL, or 0.21 m3.

Making Connections5. One consumer reaction that produces and consumes gases is the burning of propane in an outdoor barbeque. The reac-

tion is:C3H8(g) + 5 O2(g) → 3 CO2(g) + 8 H2O(g)

One industrial reaction that consumes a gas is the production of ammonium sulfate fertilizer. The reaction is:

2 NH3(g) � H2SO4(aq) → (NH4)2SO4(s)

One laboratory reaction that produces and consumes gases is the burning of methane (natural gas) in a lab burner. Thereaction is:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Reflecting6. (a) All stoichiometry involves the numerical ratios of reactants and products, that is, the ratios of amounts in moles.

(b) In an industry using a chemical reaction, knowledge of stoichiometry is essential to determine how much ofreactant to purchase and use. For example, in the reaction2 NH3(g) � H2SO4(aq) → (NH4)2SO4(s)

used to make fertilizer, the reacting amounts of ammonia and sulfuric acid would have to be calculate beforehand.(c) Adding “just the right amount” of baking powder to a baking mix is an example of consumer use of

stoichiometry.10.5 APPLICATIONS OF GASES

1 mol� CH4�6H� 2O��

0.00805 kmol� � �8.3

m1okl�P�

a��

� L� � 293 K�

��95 kPa�

132.16 g (NH4)2SO4��1 mol� (NH4)2SO4��


CHAPTER 10 SUMMARY

MAKE A SUMMARY

(Page 491)

The law is explained by kineticmolecular theory (e.g. eachmolecule has the same kineticenergy) and intermolecular forcetheory (e.g. in a gas, there arenegligible intermolecular forces).

Dalton’slaw of partial

pressures

The law provides understandingin careers such as commercialdiver, anesthetic technician,meteorologist, and welder.

Evidence and Explanation Empirical Description Applications and Careers

This law provides understandingin careers such as automobiledesigner and mechanic, chemicalengineer, welder, and barbecuedesigner.

Mixtures of gases may react(according to the law of combiningvolumes) or not (according toDalton’s law of partial pressures).

Technological applicationsexplained by this law includegasoline vapour–air mixture inautomobile engines (beforereacting) and exhaust system gasmixtures (after reaction).

Quantitative work in the laboratoryindicates that individual pressuresadd to the total measuredpressure.

Different gas molecules exert thesame pressure (because they allhave the same average kineticenergy at the same temperature).

Mixtures of gases are composedof different molecules.

This law is explained by atomic,molecular, and bonding theory.

Natural phenomena explained bythis law include Earth’s atmosphere.

The sum of individualpressures equals totalpressure.

Each gas actsindependently.

mixturesof

gases

Gases react in simplewhole-number ratiosby volume.

Natural gas, propane, andacetylene react with an optimalfuel-to-air mixture.

Molecules react in ratios thatcreate new molecules that followtheir bonding capacity.

law ofcombiningvolumes


CHAPTER 10 REVIEW

(Page 492)

Understanding Concepts1. (a) The total pressure of a mixture of nonreacting gases is equal to the sum of the partial pressures of the individual

gases in the mixture.(b) p total = p1 + p2 + p3 + …

2. Dalton’s law of partial pressures can be explained by these two concepts from the kinetic molecular theory:(i) Gas pressure is caused by the collisions of particles (molecules, atoms, ions) with the walls of the container.(ii) Gas molecules essentially act independently of each other.

Therefore, the total pressure (total of the collisions with the walls) is the sum of the individual pressures (collisions of only one kind of particle) of each gas present.

Note: Point (ii) presupposes, of course, that students know/assume that all particles at the same temperature have the same average kinetic energy.

3. p total = (230 + 13 + 7) kPa = 250 kPa4. p O2

= {100 (exactly) – 3.17} kPa = 96.83 kPa

5. (a)When measured at the same temperature and pressure, volumes of gaseous reactants and products of chemical reac-tions are found to be (to three significant digits) in simple ratios of whole numbers.

(b) 4 C7H5(NO2)3(s) � 21 O2(g) → 28 CO2(g) � 6 N2(g) � 10 H2O(g)

5.00 L


vCO2= 5.00 L � �

2281�

vCO2= 6.67 L

or

vCO2= 5.00 L� O2� � �

2

2

8

1

L

L�C

O

O

2�2�

vCO2= 6.67 L

The volume of carbon dioxide produced is 6.67 L.

vN2= 5.00 L � �

261�

vN2= 1.43 L

or

vN2= 5.00 L� O2� � �

2

6

1

L

L�N

O2

2��

vN2= 1.43 L


vH2O = 5.00 L � �1201�

vH2O = 2.38 L

or

vH2O = 5.00 L� O2� � �1

2

0

1

L

L�H

O2O

2��

vH2O = 2.38 L

The volume of water vapour produced is 2.38 L.


6. Avogadro’s idea is theoretical. It attempts to explain observed (empirical) gas behaviour by proposing that equalvolumes of gases at the same temperature and pressure contain equal numbers of molecules. This proposal is a productof intellect, not observation, and so is a theory, not a law.

7. 2 NH4NO3(s) → 2 N2(g) � 4 H2O(g) � O2(g)

1.00 mol vSATP vSATP vSATP

all gases: 24.8 L/mol (SATP)

nN2= 1.00 mol � �

22

�

nN2= 1.00 mol

vN2= 1.00 mol��

214m.8

oLl�

�

vN2= 24.8 L

or

vN2= 1.00 mol� NH4NO3� ��

2 m

2

o

m

l� N

ol�H

N

4N2�

O3��

2

1

4

m

.8

o

L

l� N

N

2�2�

vN2= 24.8 L


nH2O = 1.00 mol � �42

�

nH2O = 2.00 mol

vH2O = 2.00 mol�� 214m.8

oLl�

�

vH2O = 49.6 L

or

vH2O = 1.00 mol� NH4NO3� ��2 m

2

o

m

l�o

N

l�H

H

4

2

N

O�O�3

� � �2

1

4

m

.8

o

L

l� H

H

2

2

O�O

�

vH2O = 49.6 L


nO2= 1.00 mol � �

12

�

nO2= 0.500 mol

vO2= 0.500 mol��

214m.8

oLl�

�

vO2= 12.4 L

or

vO2= 1.00 mol� NH4NO3� � �

2 m

1

o

m

l�o

N

l�H

O

4

2�NO3�

� � �2

1

4

m

.8

o

L

l� O

O

2�2�

vO2= 12.4 L

The volume of oxygen produced is 12.4 L.The total volume of gases produced is (24.8 + 49.6 + 12.4) L = 76.8 L.

8. (a) nAr = ?

p = (100.0 + 0.400) kPa = 100.4 kPa

T = 20°C = 293 K

v = 125 mL = 0.125 L

R = 8.31 kPa � L/(mol � K)

pv = nRT


nAr = �R

pv

T�

=

nAr = 0.00515 mol = 5.15 mmol

The amount of argon gas in a bulb is 5.15 mmol.

(b) p1 = 100.4 kPa

T1 = 20°C = 293 K

p2 = ?

T2 = 200°C = 473 K

�T

p1

1� = �

T

p2

2�

p2 = �T

T2p

1

1�

=

pAr = 162 kPa

or

pAr = 100.4 kPa � �427933

K�K�

�

pAr = 162 kPa

The final argon pressure at the higher temperature will be 162 kPa.(c) nAr = ?

p = 100.4 kPa

T = 20°C = 293 K

v = 0.915 L

R = 8.31 kPa � L/(mol � K)

pv = nRT

nAr = �RpTv�

=

nAr = 0.0377 mol (in each fluorescent tube)

nAr = ?

mAr = 50 kg

MAr = 39.95 g/mol

nAr = 50.0 kg� � �319.

m95

olg�

�

nAr = 1.25 kmol (in the steel tank)

#tubes = 1.25 kmol�� 0.0

137

tu7bme

ol��

100.4 kPa� � 0.915 L��8.3

m1oklP�

a�K�� L�

� � 293 K�

473 K� � 100.4 kPa��

293 K�

100.4 kPa� � 0.125 L��8.3

m1oklP�

a�K�� L�

� � 293 K�


#tubes = 33.2 ktubes (3.32 � 104 tubes)

9. 2 ZnS(s) � 3 O2(g) → 2 ZnO(s) � 2 SO2(g)

1.00 t = 1.00 Mg vSATP

97.44 g/mol 24.8 L/mol

nZnS = 1.00 Mg� � �917.

m44

olg�

�

nZnS = 0.0103 Mmol = 10.3 kmol

nSO2= 10.3 kmol � �

22

�

nSO2= 10.3 kmol

vSO2= 10.3 kmol��

214m.8

oLl�

�

vSO2= 255 kL = 255 m3

or

vSO2= 1.00 Mg� ZnS� � �

917.

m44

ol�g�ZZnnS�S�

� � �2

2

m

m

o

o

l�l�

Z

SO

nS�2�

� � �2

1

4

m

.8

o

L

l� S

S

O

O

2�2�

vSO2= 0.255 ML = 255 kL = 255 m3

The volume of sulfur dioxide produced will be 255 kL, or 255 m3.10. 2 CaCO3(s) � 2 SO2(g) � O2(g) → 2 CaSO4(s) � 2 CO2(g)

m 500 kL (STP) vSTP

100.09 g/mol 22.4 L/mol

(a) vCO2= 500 kL ��

22

�

vCO2= 500 kL

or

vCO2= 500 kL� SO2� � �

2

2

L

L�C

SO

O

2�2�

vCO2= 500 kL

The volume of carbon dioxide gas produced at STP is 500 kL.

(b) nSO2= 500 kL� � �

212m.4

oL�l

�

nSO2= 22.3 kmol

nCaCO3= 22.3 kmol ��

22

�

nCaCO3= 22.3 kmol

mCaCO3= 22.3 kmol��

10

1

0

m

.0

o

9

l�g

�

mCaCO3= 2.23 × 103 kg = 2.23 Mg = 2.23 t

or

mCaCO3= 500 kL� SO2� � �

2

1

2.

m

4

o

L�l�S

S

O

O2�

2��

2

2

m

m

o

o

l�l�C

S

a

O

C

2�O3�

� ��10

1

0.

m

09

ol�g

C

C

a

a

C

C

O

O

3�3�

mCaCO3= 2.23 × 103 kg = 2.23 Mg = 2.23 t

The mass of calcium carbonate consumed is 2.23 t

11. SO3(g) � H2O(l) → H2SO4(aq)

1.00 t = 1.00 Mg v


80.06 g/mol 0.12 mmol/L

nSO3= 1.00 Mg� � �

810.

m06

olg�

�

nSO3= 0.0125 Mmol

nH2SO4= 0.0125 Mmol � �

11

�

nH2SO4= 0.0125 Mmol

vH2SO4= 0.0125 Mmol��

0.121

mL

mol��

vH2SO4= 1.0 × 108 L = 0.10 GL

or

vH2SO4= 1.00 Mg� SO3� � �

8

1

0.

m

06

ol�g�S

S

O

O3�

3��

1

1

m

m

o

o

l�l�H

S2

O

S

3�O4�

� ��0.12

1

m

L

m

H

o2S

l�O

H4

2SO4��

vH2SO4= 1.0 × 108 L = 0.10 GL

The volume of sulfuric acid that could be formed is 0.10 GL.12. 2 H2O(l) → 2 H2(g) � O2(g)

50.0 mL v

18.02 g/mol

Both gases measured at 23°C = 296 K and 103 kPa.

(a) vO2= 50.0 mL � �

12

�

vO2= 25.0 mL

or

vO2= 50.0 mL� H2� � �

2

1

L�L O

H2

2��

vO2= 25.0 mL

The volume of oxygen gas produced is 25.0 mL.(b) p1 = 103 kPa

T1 = 23°C = 296 K

v1 = 50.0 mL

p2 = 101.325 kPa

T2 = 0°C = 273 K

vH2= ?

�p

T1v

1

1� = �p

T2v

2

2�

vH2= �

T

T2p

1p1v

2

1�

=

vH2= 46.9 mL

or

vH2= 50.0 mL � �

10110.332k5Pka�Pa�

� � �227936

K�K�

�

273 K� � 103 kPa� � 50.0 mL��

296 K� � 101.325 kPa�


vH2= 46.9 mL

The volume of the hydrogen at STP would be 46.9 mL.

(c) Note: The mass of water may be calculated at this point in this question from the volume of either hydrogen oroxygen, and if using hydrogen values, from the volume at either set of pressure and temperature conditions. For best accuracy, and to avoid perpetuating possible errors in preceding steps, calculating from original values is usu-ally preferable, where possible.

nH2= ?

p = 103 kPa

T = 23°C = 296 K

v = 50.0 mL

R = 8.31 kPa � L/(mol � K)

pv = nRT

nH2= �

Rpv

T�

=

nH2= 2.09 mmol

nH2O = 2.09 mmol � �22

�

nH2O = 2.09 mmol

mH2O = 2.09 mmol�� 118.

m02

ol�g

�mH2O = 37.7 mg

or

mH2O = 50.0 mL� H2��8.3

1

1

m

kP

o

a�l� H

� 12� �

L�K�

H2��

120936kK�Pa�

� � �2

2

m

m

o

o

l�l�H

H2

2�O�

� � �1

1

8.

m

02

ol�g

H

H

2

2

O�O

�

mH2O = 37.7 mg

The mass of water decomposed is 37.7 mg.13. According to Dalton’s law of partial pressures and Avogadro’s theory, the partial pressure of any gas in a mixture must

be proportional to its mole fraction. Since the mixture has 2.00 mol N2(g) and 3.00 mol H2(g), totalling 5.00 mol; andthe total pressure is 200 kPa;

pH2= 200 kPa � �

35..0000

mm

ool�l�

�

pH2= 120 kPa

pN2= ptotal – pH2

= (200 – 120) kPa = 80 kPa

The partial pressure of nitrogen is 80 kPa; and the partial pressure of hydrogen is 120 kPa. 14. a) When temperature increases, the molar volume of a gas increases.

b) When pressure increases, the molar volume of a gas decreases.


The molar volume of propane at STP will be 22.4 L/mol, just as for any other gas, according to Avogadro’s theory.(Assume propane behaves like the ideal gas.)

103 kPa� � 50.0 mL��8.3

m1oklP�

a�K�� L�

� � 296 K�


(b) AnalysismC3H8

= (426.79 – 424.92) g = 1.87 gp1 = (98.23 – 2.49) kPa = 95.74 kPav1 = 1065 mL

T1 = 21.0°C = 294 K

p2 = 101.325 kPa

vC3H8= ?

T2 = 0°C = 273 K

MC3H8= 44.11 g/mol

VSTP = ?

nC3H8= 1.87 g� � �

414.

m11

olg�

�

nC3H8= 0.0424 mol = 42.4 mmol

�p

T1v

1

1� = �p

T2v

2

2�

vC3H8=

=

vC3H8= 934 mL

or

vC3H8= 1065 mL � �

19051..73425

kPkP

a�a�

� �

vC3H8= 934 mL

The volume of propane at STP would be 934 mL. Since this is an amount of 42.4 mmol, the molar volume can nowbe calculated as follows:

VSTP = �42

9.344m�m�mLol

�

VSTP = 22.0 L/mol

According to the evidence from this investigation, the molar volume of propane at STP is 22.0 L/mol.

(c) Evaluationdifference = 22.4 – 22.0 L/mol = 0.4 L/mol

% difference = �202.4.4

LL//mm

ool�l�

� � 100% = 2%

The prediction is judged to be verified, because the value calculated from the evidence is in close agreement with it,to within 2%.

(d) EvaluationAvogadro’s theory is supported by the result of this investigation, because the result agrees well with the predictionmade from this authority.

16. Experimental DesignA cylinder of compressed gas is used to release a noble gas, which is collected at ambient conditions by water dis-placement. The amount is calculated from volume, temperature, and pressure measurements. The molar mass is cal-culated from the amount and the measured mass, and is used to identify the gas.

273 K��

273 K�� 95.74 kPa� � 1065 mL��

294 K�� 101.325 kPa�

T2p1v1�


17. One natural and one technological use or source for each of the following gases:(a) oxygen from plant respiration for welding(b) methane from plant decomposition for fuel(c) helium from natural gas wells for balloons(d) air Earth’s atmosphere source of gases(e) water vapour from animal respiration for humidifiers(f) carbon dioxide from combustion fire extinguishers

Making Connections18. (a) Freons were initially used as refrigerants, and soon became popular as aerosol propellants, and then later as the

foaming agent for making foam plastics, and as a non-stick, non-toxic solvent for use by the electronics industry.(b) The production of Freons is banned in many countries, including Canada, because these compounds react in the

upper atmosphere to produce chlorine atoms — which, in turn, catalyze the decomposition of ozone. Thisincreases environmental damage from ultraviolet rays.

(c) mCF2Cl2= 1.00 kg

MCF2Cl2= 120.91 g/mol

VCF2Cl2= 24.8 L/mol (SATP)

nCF2Cl2= 1.00 kg� � �

1210m.9

o1lg�

�

nCF2Cl2= 0.00827 kmol = 8.27 mol

vCF2Cl2= 8.27 mol��

214m.8

oLl�

�

vCF2Cl2= 205 L

or

vCF2Cl2= 1.00 kg� � �

1210m.9

o1l�g�

� � �214m.8

oLl�

�

vCF2Cl2= 0.205 kL = 205 L

The volume of Freon at SATP would be 205 L.

19. 2 C2H5SH(g) � 9 O2(g) → 4 CO2(g) � 6 H2O(g) � 2 SO2(g)

1.00 g v v v

62.14 g/mol All gases: 24.8 L/mol (SATP)

nC2H5SH = 1.00 g� � �612.

m14

olg�

�

nC2H5SH = 0.0161 mol = 16.1 mmol

nCO2= 16.1 mmol � �

42

�

nCO2= 32.2 mmol

vCO2= 32.2 mmol��

214m.8

oLl�

�

vCO2= 798 mL

or

vCO2= 1.00 g� C2H5SH� ��

6

1

2.

m

14

ol�g�C

C2

2

H

H5

5

S

S

H�H�

� � �2

4

m

m

o

o

l�l�C

C

2H

O

5

2�SH�

� � �2

1

4

m

.8

o

L

l� C

C

O

O

2�2�

vCO2= 0.798 L = 798 mL

The volume of carbon dioxide produced is 798 mL.

nH2O = 16.1 mmol � �62

�


nH2O = 48.3 mmol

or vH2O= 48.3 mmol� � �

214m.8

oLl�

�

vH2O= 1.20 � 103 mL = 1.20 L

vH2O= 1.00 g� C2H5SH� ��

6

1

2.

m

14

ol�g�C

C2

2

H

H5

5

S

S

H�H�

� ��2

6

m

m

ol�o

C

l�

2

H

H2

5

O�SH�

� � �2

1

4

m

.8

o

L

l� H

H

2

2

O�O

�

vH2O = 1.20 L


nSO2= 16.1 mmol � �

22

�nSO2

= 16.1 mmol

vSO2= 16.1 mmol� � �

214m.8

oLl�

�

vSO2= 399 mL

or

vSO2 = 1.00 g� C2H5SH� ��

6

1

2.

m

14

ol�g�C

C2

2

H

H5

5

S

S

H�H�

� ��2 m

2 m

ol�o

C

l�

2

S

H

O

5

2�SH�

� � �2

1

4

m

.8

o

L

l� S

S

O

O

2�2�

vSO2= 0.399 L = 399 mL

The volume of sulfur dioxide produced is 399 mL.

(b) Ethanethiol added to natural gas increases safety when the natural gas is used as a heating fuel, since any leak can bedetected quickly, before the area becomes an explosion hazard. However, the compound is hazardous to handle inundiluted form before it is mixed with natural gas, and it produces the pollutant sulfur dioxide when burned.

20. A typical answer might include discussion of the production of nitrogen oxides, symbolized NOX, by heavy vehicletraffic in metropolitan areas. NOX encourages smog formation, reacting with sunlight and oxygen to cause an increasein ground-level ozone. Ozone, O3(g), is thought to be the primary agent causing smog damage to vegetation, and is atoxic and irritant gas to humans, interfering with lung function.

UNIT 4 PERFORMANCE TASK

ReportThe report will require extensive student research, the results of which cannot be simulated here. The report should cover

the main points:• historical development of the technology;

• scientific principles used in the technology, especially those relating to gases;

• risks and benefits (to society and the environment) of use of the technology;

• a summary of a related career.

UNIT 4 REVIEW

(Page 496)

Understanding Concepts1. (a) The intermolecular forces in solid carbon dioxide must be very weak, because the molecules separate completely

at very low temperature.


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CHAPTER 10 GAS MIXTURES AND REACTIONS - Quia