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C H A P T E R 12
THE GASEOUS STATE OF MATTER
SOLUTIONS TO REVIEW QUESTIONS
1. The pressure of a gas is the force that gas particles exert on the walls of a container. The pressure
depends on the temperature, the number of molecules of the gas and the volume of the container.
2. The air pressure inside the balloon is greater than the air pressure outside the balloon. The pressure
inside must equal the sum of the outside air pressure plus the pressure exerted by the stretched rubber of
the balloon.
3. The major components of dry air are nitrogen and oxygen.
4. 1 torr ¼ 1 mm Hg
5. The molecules of H2 at 100 C are moving faster. Temperature is a measure of average kinetic energy.
At higher temperatures, the molecules will have more kinetic energy.
6. 1 atm corresponds to 4 L.
7. The pressure times the volume at any point on the curve is equal to the same value. This is an inverse
relationship as is Boyle’s law. (PV ¼ k)
8. If T2 < T1, the volume of the cylinder would decrease (the piston would move downward).
! 2 NOðgÞ
9. N2 ðgÞ þ O2 ðgÞ 1 vol þ 1 vol ! 2 vol
According to Avogadro’s law, equal volumes of nitrogen and oxygen at the same temperature and
pressure contain the same number of molecules. In the reaction, nitrogen and oxygen molecules react in
a 1:1 ratio. Since two volumes of nitrogen monoxide are produced, one molecule of nitrogen and one
molecule of oxygen must produce two molecules of nitrogen monoxide. Therefore each nitrogen and
oxygen molecule must be made up of two atoms (diatomic).
10. We refer gases to STP because some reference point is needed to relate volume to moles. A temperature
and pressure must be specified to determine the moles of gas in a given volume, and 0 C and 760 torr
are convenient reference points.
11. Gases are described by the following parameters:
(a) pressure
(c) temperature
(b) volume
(d) number of moles
12. An ideal gas is one which follows the described gas laws at all P, V, and T and whose behavior is
described exactly by the Kinetic Molecular Theory.
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- Chapter 12 13. Boyle’s law: P1 V1 ¼ P2 V2 , ideal gas equation: PV ¼ nRT
If you have an equal number of moles of two gases at the same temperature the right side of the ideal
gas equation will be the same for both gases. You can then set PV for the first gas equal to PV for the
second gas (Boyle’s law) because the right side of both equations will cancel.
14. Charles’ law: V1 =T1 ¼ V2 =T2 , ideal gas equation: PV ¼ nRT
Rearrange the ideal gas equation to: V=T ¼ nR=P
If you have an equal number of moles of two gases at the same pressure the right side of the rearranged
ideal gas equation will be the same for both. You can set V=T for the first gas equal to V=T for the
second gas (Charles’ law) because the right side of both equations will cancel.
15. Basic assumptions of Kinetic Molecular Theory include:
(a) Gases consist of tiny particles.
(b) The distance between particles is great compared to the size of the particles.
(c) Gas particles move in straight lines. They collide with one another and with the walls of the
container with no loss of energy.
(d) Gas particles have no attraction for each other.
(e) The average kinetic energy of all gases is the same at any given temperature. It varies directly
with temperature.
16. The order of increasing molecular velocities is the order of decreasing molar masses.
increasing molecular velocity
!
Rn; F2 ; N2 CH4 ; He; H2
!
decreasing molar mass
At the same temperature the kinetic energies of the gases are the same and equal to ½ mv2 . For the
kinetic energies to be the same, the velocities must increase as the molar masses decrease.
17. Average kinetic energies of all these gases are the same, since the gases are all at the same temperature.
18. A gas is least likely to behave ideally at low temperatures. Under this condition, the velocities of the
molecules decrease and attractive forces between the molecules begin to play a significant role.
19. A gas is least likely to behave ideally at high pressures. Under this condition, the molecules are forced
close enough to each other so that their volume is no longer small compared to the volume of the
container. Attractive forces may also occur here and sooner or later, the gas will liquefy.
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- Chapter 12 20. Behavior of gases as described by the Kinetic Molecular Theory.
(a) Boyle’s law. Boyle’s law states that the volume of a fixed mass of gas is inversely proportional to
the pressure, at constant temperature. The Kinetic Molecular Theory assumes the volume
occupied by gases is mostly empty space. Decreasing the volume of a gas by compressing it,
increases the concentration of gas molecules, resulting in more collisions of the molecules and
thus increased pressure upon the walls of the container.
(b) Charles’ law. Charles’ law states that the volume of a fixed mass of gas is directly proportional to
the absolute temperature, at constant pressure. According to Kinetic Molecular Theory, the kinetic
energies of gas molecules are proportional to the absolute temperature. Increasing the temperature
of a gas causes the molecules to move faster, and in order for the pressure not to increase, the
volume of the gas must increase.
(c) Dalton’s law. Dalton’s law states that the pressure of a mixture of gases is the sum of the pressures
exerted by the individual gases. According to the Kinetic Molecular Theory, there are no attractive
forces between gas molecules; therefore, in a mixture of gases, each gas acts independently and
the total pressure exerted will be the sum of the pressures exerted by the individual gases.
21. Conversion of oxygen to ozone is an endothermic reaction. Evidence for this statement is that energy
(286 kJ=3 mol O2) is required to convert O2 to O3.
22.
Oxygen atom ¼ O Oxygen molecule ¼ O2
An oxygen molecule contains 16 electrons.
Ozone molecule ¼ O3
23. The pressure inside the bottle is less than atmospheric pressure. We come to this conclusion because the
water inside the bottle is higher than the water in the trough (outside the bottle).
24. The density of air is 1.29 g/L. Any gas listed below air in Table 12.4 has a density greater than air. For
example: O2, H2S, HCl, F2, CO2.
25. Equal volumes of H2 and O2 at the same T and P:
(a) have equal number of molecules (Avogadro’s law)
(b) mass O2 ¼ 16 times mass of H2
(c) moles O2 ¼ moles H2
(d) average kinetic energies are the same (T same)
(e) density O2 ¼ 16 times the density of H2
mass O2
mass H2
density O2 ¼
density H2 ¼
volume O2
volume H2
volume O2 ¼ volume H2
mass O2
mass H2
mass O2
ðdensity H2 Þ
density O2 ¼
density O2
density H2
mass H2
32
ðdensity H2 Þ ¼ 16 ðdensity H2 Þ
density O2 ¼
2
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- Chapter 12 26. Heating a mole of N2 gas at constant pressure has the following effects:
(a) Density will decrease. Heating the gas at constant pressure will increase its volume. The mass
does not change, so the increased volume results in a lower density.
(b) Mass does not change. Heating a substance does not change its mass.
(c) Average kinetic energy of the molecules increases. This is a basic assumption of the Kinetic
Molecular Theory.
(d) Average velocity of the molecules will increase. Increasing the temperature increases the average
kinetic energies of the molecules; hence, the average velocity of the molecules will increase also.
(e) Number of N2 molecules remains unchanged. Heating does not alter the number of molecules
present, except if extremely high temperatures were attained. Then, the N2 molecules might
dissociate into N atoms resulting in fewer N2 molecules.
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- Chapter 12 -
SOLUTIONS TO EXERCISES
1. Pressure conversions:
a.
b.
c.
(a)
(b)
(c)
torr
inches Hg
768
752
745
30.2
29.6
29.3
kilopascals
102
100.
99.3
760 torr
¼ 768 torr
in: Hg ! torr ð30:2 in: HgÞ
29:9 in: Hg
101:325 kPa
in: Hg ! kPa ð30:2 in: HgÞ
¼ 102 kPa
29:9 in: Hg
29:9 in: Hg
¼ 29:6 in: Hg
torr ! in: Hg ð752 torrÞ
760 torr
101:325 kPa
torr ! kPa ð752 torrÞ
¼ 100: kPa
760 torr
760 torr
¼ 745 torr
kPa ! torr ð99:3 kPaÞ
101:325 kPa
29:9 in: Hg
kPa ! in: Hg ð99:3 kPaÞ
¼ 29:3 in: Hg
101:325 kPa
2. Pressure conversions:
a.
b.
c.
(a)
(b)
(c)
mm Hg
lb/in.2
atmospheres
789
1700
1100
15.3
32
21
1.04
2.2
1.4
14:7 lb=in:2
¼ 15:3 lb=in:2
mm Hg ! lb=in: ð789 mm HgÞ
760 mm Hg
1 atm
mm Hg ! atm ð789 mm HgÞ
¼ 1:04 atm
760 mm Hg
760 mm Hg
2
2
¼ 1700 mm Hg
lb=in: ! mm Hg ð32 lb=in: Þ
14:7 lb=in:2
1 atm
lb=in:2 ! atm ð32 lb=in:2 Þ
¼ 2:2 atm
14:7 lb=in:2
760 mm Hg
atm ! mm Hg ð1:4 atmÞ
¼ 1100 mm Hg
1 atm
14:7 lb=in:2
2
atm ! lb=in: ð1:4 atmÞ
¼ 21 lb=in:2
1 atm
2
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- Chapter 12 3. (a)
(b)
(c)
(d)
(e)
4. (a)
(b)
(c)
(d)
(e)
5. (a)
(b)
(c)
6. (a)
101:3 kPa
torr ! kPa ð953 torrÞ
¼ 127 kPa
760 torr
1 atm
¼ 0:0294 atm
kPa ! atm ð2:98 kPaÞ
101:3 kPa
760 mm Hg
¼ 2110 mm Hg
atm ! mm Hg ð2:77 atmÞ
1 atm
1 atm
¼ 0:489 atm
torr ! atm ð372 torrÞ
760 torr
76:0 cm Hg
¼ 214 cm Hg
atm ! cm Hg ð2:81 atmÞ
1 atm
101:3 kPa
torr ! kPa ð649 torrÞ
¼ 86:5 kPa
760 torr
1 atm
¼ 0:0500 atm
kPa ! atm ð5:07 kPaÞ
101:3 kPa
760 mm Hg
¼ 2770 mm Hg
atm ! mm Hg ð3:64 atmÞ
1 atm
1 atm
¼ 1:06 atm
torr ! atm ð803 torrÞ
760 torr
76:0 cm Hg
¼ 82:1 cm Hg
atm ! cm Hg ð1:08 atmÞ
1 atm
1 atm
2
2
lb=in: ! atm ð1920 lb=in: Þ
¼ 131 atm
14:7 lb=in:2
760 torr
2
2
lb=in: ! torr ð1920 lb=in: Þ
¼ 9:93 104 torr
14:7 lb=in:2
101:3 kPa
2
2
lb=in: ! kPa ð1920 lb=in: Þ
¼ 1:32 104 kPa
14:7 lb=in:2
lb=in: ! atm ð31 lb=in: Þ
2
2
1 atm
14:7 lb=in:2
¼ 2:1 atm
(b)
760 torr
¼ 1600 torr
lb=in: ! torr ð31 lb=in: Þ
14:7 lb=in:2
(c)
101:3 kPa
lb=in: ! kPa ð31 lb=in: Þ
¼ 210 kPa
14:7 lb=in:2
2
2
2
2
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- Chapter 12 P 1 V1
V2
ð825 torrÞð725 mLÞ
¼ 2110 torr
ð283 mLÞ
ð825 torrÞð725 mLÞ
Change 2:87 L to mL
¼ 208 torr
ð2:87 LÞð1000 mL=1 LÞ
7. P1 V1 ¼ P2 V2 or P2 ¼
(a)
(b)
P 1 V1
V2
ð508 torrÞð486 mLÞ
¼ 1330 torr
ð185 mLÞ
ð508 torrÞð486 mLÞ
Change 6:17 L to mL
¼ 40:0 torr
ð6:17 LÞð1000 mL=1 LÞ
8. P1 V1 ¼ P2 V2 or P2 ¼
(a)
(b)
9. P1 V1 ¼ P2 V2 or V2 ¼
P 1 V1
P2
ð58:2 LÞð7:25 atmÞ
¼ 208 L
ð2:03 atmÞ
10. P1 V1 ¼ P2 V2 or V2 ¼
P 1 V1
P2
ð832 LÞð0:204 atmÞ
¼ 21:2 L
ð8:02 atmÞ
11.
V1 V2
¼
T1 T2
(a)
(b)
(c)
12.
(b)
(c)
V2 ¼
V1 T 2
; temperatures must be in Kelvin ( C þ 273)
T1
ð125 mLÞð268 KÞ
¼ 114 mL
294 K
ð125 mLÞð308 KÞ
¼ 131 mL
294 K
ð125 mLÞð1095 KÞ
¼ 466 mL
294 K
V1 V2
¼
T1 T2
(a)
or
or
V2 ¼
V1 T 2
; temperatures must be in Kelvin ( C þ 273)
T1
ð575 mLÞð298 KÞ
¼ 691 mL
248 K
ð575 mLÞð273 KÞ
¼ 633 mL ð32 F ¼ 0 C ¼ 273 KÞ
248 K
ð575 mLÞð318 KÞ
¼ 737 mL
248 K
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- Chapter 12 -
13. Use the combined gas laws
V2 ¼
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
P1 V1 T2
P2 T1
P 1 V1 P 2 V2
¼
T1
T2
or V2 ¼
P 1 V1 T 2
P2 T1
ð0:950 atmÞð1400: LÞð275 KÞ
¼ 2:4 105 L
ð4:0 torrÞð1 atm=760 torrÞð291 KÞ
16. Use the combined gas law
V2 ¼
P1 V1 T2
P2 T1
ð678 torrÞð25:6 LÞð308 KÞ
¼ 30:8 L
ð595 torrÞð292 KÞ
15. Use the combined gas law
V2 ¼
or V2 ¼
ð0:75 atmÞð1025 mLÞð308 KÞ
¼ 544 mL
ð1:25 atmÞð348 KÞ
14. Use the combined gas laws
V2 ¼
P1 V1 P2 V2
¼
T1
T2
P 1 V1 P 2 V2
¼
T1
T2
or V2 ¼
P 1 V1 T 2
P2 T1
or P2 ¼
P 1 V1 T 2
V2 T1
or T2 ¼
P 2 V2 T 1
P1 V1
ð2:50 atmÞð22:4 LÞð268 KÞ
¼ 33:4 L
ð1:50 atmÞð300: KÞ
17. Use the combined gas law
P 1 V1 P 2 V2
¼
T1
T2
ð1:0 atmÞð775 mLÞð298 KÞ
¼ 1:4 atm
ð615 mLÞð273 KÞ
18. Use the combined gas law
P 1 V1 P 2 V2
¼
T1
T2
Change 765 torr to atmospheres.
ð765 torrÞð1 atm=760 torrÞð1:5 LÞð292 KÞ
¼ 120 K ð120 K 273Þ; K ¼ 153 C
ð1:5 atmÞð2:5 LÞ
19. Ptotal ¼ PO2 þ PH2 O vapor ¼ 772 torr
PH2 O vapor ¼ 21:2 torr
PO2 ¼ 772 torr 21:2 torr ¼ 751 torr
20. Ptotal ¼ PCH4 þ PH2 O vapor ¼ 749 mm Hg
PH2 O ¼ 30:0 torr ¼ 30:0 mm Hg
PCH4 ¼ 749 mm Hg 30:0 mm Hg ¼ 719 mm Hg
21.
Ptotal ¼ PN2 þ PH2 þ PO2
¼ 200: torr þ 600: torr þ 300: torr ¼ 1100: torr ¼ 1:100 103 torr
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- Chapter 12 22.
Ptotal ¼ PH2 þ PN2 þ PO2
¼ 325 torr þ 475 torr þ 650: torr ¼ 1450: torr ¼ 1:450 103 torr
23. Ptotal ¼ PCH4 þ PH2 O vapor (Solubility of methane is being ignored.)
PH2 O vapor ¼ 23:8 torr
PCH4 ¼ 720: torr 23:8 torr ¼ 696 torr
To calculate the volume of dry methane, note that the temperature is constant, so P1 V1 ¼ P2 V2 can be used.
V2 ¼
24.
P1 V1 ð696 torrÞð2:50 L CH4 Þ
¼ 2:29 L CH4
¼
ð760: torrÞ
P2
Ptotal ¼ PC3 H8 þ PH2 O vapor
C3 H8 is propane
PH2 O vapor ¼ 20:5 torr
PC3 H8 ¼ 745 torr 20:5 torr ¼ 725 torr
To calculate the volume of dry propane, note that the temperature is constant, so P1 V1 ¼ P2 V2
can be used.
V2 ¼
P1 V1 ð725 torrÞð1:25 L C3 H8 Þ
¼
¼ 1:19 L C3 H8
ð760: torrÞ
P2
25. 1 mol of a gas occupies 22.4 L at STP
22:4 L
ð6:26 mol N2 Þ
¼ 140: L N2
1 mol
26. 1 mol of a gas occupies 22.4 L at STP
22:4 L
ð5:89 mol CO2 Þ
¼ 132 L CO2
1 mol
27. (a)
(b)
(c)
28. (a)
(b)
(c)
22:4 L
¼ 22:4 L CO2
6:022 1023 molecules
6:022 10 molecules CO2
22:4 L
ð2:5 mol CH4 Þ
¼ 56 L CH4
mol
22:4 L
ð12:5 g O2 Þ
¼ 8:75 L O2
32:00 g
22:4 L
24
1:80 10 molecules SO3
¼ 67:0 L SO3
6:022 1023 molecules
22:4 L
ð7:5 mol C2 H6 Þ
¼ 170 L C2 H6
mol
22:4 L
ð25:2 g Cl2 Þ
¼ 7:96 L Cl2
70:90 g
23
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- Chapter 12 29. ð725 mL NH3 Þ
1L
1 mol
17:03 g
¼ 0:551 g NH3
1000 mL 22:4 L
mol
30. ð945 mL C3 H6 Þ
1L
1 mol
42:08 g
¼ 1:78 g C3 H6
1000 mL 22:4 L
mol
1 mol
22:4 L
31. ð1025 molecules CO2 Þ
¼ 3:813 1020 L CO2
mol
6:022 1023 molecules
1 mol
6:022 1023 molecules
32. ð10:5 L CO2 Þ
¼ 2:82 1023 molecules CO2
22:4 L
mol
33. density of Cl2 gas ¼ 3:17 g=L (from table 12.4)
1L
ð10:0 g Cl2 Þ
¼ 3:15 L Cl2
3:17 g
34. density of CH4 gas ¼ 0:716 g=L (from table 12.4)
ð3:0 L CH4 Þð0:716 g=LÞ ¼ 2:1 g CH4
35. PV ¼ nRT V ¼
nRT
P
V¼
36. PV ¼ nRT V ¼
nRT
P
V¼
37. PV ¼ nRT n ¼
PV
RT
n¼
ð75 mol NH3 Þð0:0821 L atm=mol KÞð295 KÞ
¼ 1:9 103 L NH3
729 torr
torr
760
atm
ð105 mol CH4 Þð0:0821 L atm=mol KÞð312 KÞ
¼ 1:8 103 L CH4
1:5 atm
ð1:2 atmÞð5:25 L O2 Þ
¼ 0:26 mol O2
ð0:0821 L atm=mol KÞð299 KÞ
1
0
38. PV ¼ nRT n ¼
PV
RT
B752 torrC
@
torr Að9:55 L CO2 Þ
760
atm
n¼
¼ 0:362 mol CO2
ð0:0821 L atm=mol KÞð318 KÞ
0
39. PV ¼ nRT T ¼
PV
nR
B732 torrC
@
torr Að645 L XeÞ
760
atm
¼ 300: K
T¼
ð25:2 mol XeÞð0:0821 L atm=mol KÞ
0
40. PV ¼ nRT T ¼
PV
nR
1
1
B675 torrC
@
torr Að725 L ArÞ
760
atm
T¼
¼ 209 K
ð37:5 mol ArÞð0:0821 L atm=mol KÞ
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- Chapter 12 41. (a)
(b)
(c)
(d)
42. (a)
(b)
(c)
(d)
43. (a)
Density
of Gases 4:003 g He
1 mol
¼ 0:179 g=L He
d¼
mol
22:4 L
20:01 g HF
1 mol
d¼
¼ 0:893 g=L HF
mol
22:4 L
42:08 g C3 H6
1 mol
d¼
¼ 1:89 g=L C3 H6
22:4 L
mol
120:9 g CCl2 F2
1 mol
d¼
¼ 5:40 g=L CCl2 F2
22:4 L
mol
222 g Rn
1 mol
d¼
¼ 9:91 g=L Rn
mol
22:4 L
46:01 g NO2
1 mol
d¼
¼ 2:05 g=L NO2
22:4 L
mol
80:07 g SO3
1 mol
d¼
¼ 3:57 g=L SO3
22:4 L
mol
28:05 g C2 H4
1 mol
d¼
¼ 1:25 g=L C2 H4
22:4 L
mol
Assume 1.00 mol of NH3 and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol NH3 Þð0:0821 L atm=mol KÞð298 KÞ
¼
¼ 20: L NH3 at 25 C and 1:2 atm
P
1:2 atm
17:03 g
¼ 0:85 g=L NH3
d¼
20: L NH3
V¼
(b)
Assume 1.00 mol of Ar and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol ArÞð0:0821 L atm=mol KÞð348 KÞ
¼
¼ 29:1 L Ar at 75 C and 745 torr
745 torr
P
torr
760
39:95 g
atm
¼ 1:37 g=L Ar
d¼
29:1 L Ar
V¼
44. (a)
Assume 1.00 mol C2H4 and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol C2 H4 Þð0:0821 L atm=mol KÞð305 KÞ
¼
¼ 33 L C2 H4 at 32 C and 0:75 atm
P
0:75 atm
28:05 g
¼ 0:85 g=L C2 H4
d¼
33 L C2 H4
Assume 1.00 mol of He and determine the volume using the ideal gas equation, PV ¼ nRT.
V¼
(b)
nRT ð1:00 mol HeÞð0:0821 L atm=mol KÞð330: KÞ
¼ 26:0 L He at 57 C and 791 torr
¼
791 torr
P
torr
760
atm
4:003 g
¼ 0:154 g=L He
d¼
26:0 L He
V¼
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- Chapter 12 45. The balanced equation is CaCO3(s) ! CaO(s) + CO2(g); 1 mol of a gas occupies 22.4 L at STP.
1 mol
1 mol CO2
22:4 L CO2
(a) ð6:24 g CaCO3 Þ
¼ 1:40 L CO2
100:1 g 1 mol CaCO3
1 mol CO2
1 mol CO2
1 mol CaCO3
100:1 g
(b) ð52:6 L CO2 Þ
¼ 235 g CaCO3
1 mol
22:4 L CO2
1 mol CO2
46. The balanced equation is Mg(s) + 2 HCl(aq) ! MgCl2(aq) + H2(g); 1 mol of a gas occupies 22.4 L
at STP.
1 mol
1 mol H2
22:4 L 1000 mL
¼ 3:95 104 mL H2
(a) ð42:9 g MgÞ
24:31
g
1
mol
1
L
1
mol
Mg
1L
1 mol H2
2 mol HCl
(b) ð825 mL H2 Þ
¼ 0:0737 mol HCl
1000 mL 22:4 L H2
1 mol H2
47. The balanced equation is 4 NH3 ðgÞ þ 5 O2 ðgÞ ! 4 NOðgÞ þ 6 H2 OðgÞ
Remember that volume–volume relationships are the same as mole–mole relationships when dealing
with gases at the same T and P.
5 L O2
¼ 3:1 L O2
(a) ð2:5 L NH3 Þ
4 L NH3
6 L H2 O
1 mol
18:02 g
(b) ð25 L NH3 Þ
¼ 30: g H2 O
4 L NH3
22:4 L
1 mol
(c)
Limiting reactant problem.
4 L NO
ð25 L O2 Þ
¼ 20: L NO
5 L O2
4 L NO
ð25 L NH3 Þ
¼ 25 L NO
4 L NH3
Oxygen is the limiting reactant. 20. L NO is formed.
48. The balanced equation is C3 H8 ðgÞ þ 5 O2 ðgÞ ! 3 CO2 ðgÞ þ 4 H2 OðgÞ
Remember that volume–volume relationships are the same as mole–mole relationships when dealing
with gases at the same T and P.
5 L O2
(a) ð7:2 L C3 H8 Þ
¼ 36 L O2
1 L C3 H8
3 L CO2
1 mol
44:01 g
(b) ð35 L C3 H8 Þ
¼ 210 g CO2
22:4 L
1 mol
1 L C3 H8
(c)
Limiting reactant problem.
4 L H2 O
ð15 L C3 H8 Þ
¼ 60: L H2 O
1 L C3 H8
4 L H2 O
¼ 12 L H2 O
ð15 L O2 Þ
5 L O2
Oxygen is the limiting reactant. 12 L H2O is formed.
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- Chapter 12 49. The balanced equation is 2 KClO3 ðsÞ ! 2 KClðsÞ þ 3 O2 ðgÞ
1000 g
1 mol
3 mol O2
22:4 L
¼ 237 L O2
ð0:525 kg KClÞ
1 kg
74:55 g 2 mol KCl
1 mol
50. The balanced equation is C6 H12 O6 ðsÞ þ 6 O2 ðgÞ ! 6 CO2 ðgÞ þ 6 H2 Oðl Þ
1000 g
1 mol
6 mol CO2
22:4 L
ð1:50 kg C6 H12 O6 Þ
¼ 1:12 103 L CO2
1 kg
180:2 g 1 mol C6 H12 O6
1 mol
51. Like any other gas, water in the gaseous state occupies a much larger volume than in the liquid state.
52. During the winter the air in a car’s tires is colder, the molecules move slower and the pressure
decreases. In order to keep the pressure at the manufacturer’s recommended psi air needs to be added to
the tire. The opposite is true during the summer.
53. Sample (b) has the greatest pressure because it has more molecules. Pressure is proportional to number
of moles when temperature and volume remain constant.
54. Image (a) best represents the balloon because lowering the temperature of the gas will decrease its
volume. The size of the balloon will also decrease, but the gas molecules will still be distributed
throughout the volume of the balloon.
55. (a)
(b)
(c)
(d)
the pressure will be cut in half
the pressure will double
the pressure will be cut in half
the pressure will increase to 3.7 atm or 2836 torr
nRT
V
ð1:5 molÞð0:0821 L atm=mol KÞð303 KÞ
P¼
¼ 3:7 atm
10: L
760 torr
P ¼ 3:7 atm
¼ 2:8 103 torr
1 atm
PV ¼ nRT P ¼
56. (a)
(c)
P
n
V
V
(b)
T
V
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- Chapter 12 57.
Law
Boyle’s law
Charles’ law
Avogadro’s law
Graph showing
the relationship
of variable factors
Factors that
are constant
Factors that
are variable
Temperature and number
of moles
Pressure and number of moles
Pressure and temperature
Pressure and volume
(c)
Volume and temperature
Volume and number of moles
(b)
(b)
58. The can is a sealed unit and very likely still contains some of the aerosol. As the can is heated, pressure
builds up in it eventually causing the can to explode and rupture with possible harm from flying debris.
59. One mole of an ideal gas occupies 22.4 liters at standard conditions. (0 C and 1 atm pressure)
PV ¼ nRT
ð1:00 atmÞðVÞ ¼ ð1:00 molÞð0:0821 L atm=mol KÞð273 KÞ
V ¼ 22:4 L
60. Solve for volume using PV ¼ nRT
(a)
V¼
ð0:2 mol Cl2 Þð0:0821 L atm=mol KÞð321 KÞ
¼ 5 L Cl2
ð80 cm=76 cmÞ atm
1 mol
0:0821 L atm
ð4:2 g NH3 Þ
ð262 KÞ
17:03 g
mol K
(b) V ¼
¼ 8:2 L NH3
0:65 atm
1 mol
0:0821 L atm
ð21 g SO3 Þ
ð328 KÞ
80:07 g
mol K
(c) V ¼
¼ 6:5 L SO3
110 kPa
kPa
101:3
atm
4.2 g NH3 has the greatest volume
61. (a)
(b)
1 mol of a gas occupies 22.4 L at STP
1 mol CH4
6:022 1023 molecule CH4
ð1 L CH4 Þ
¼ 2:69 1022 molecules CH4
22:4 L CH4
1 mol CH4
Temperature must be in K; must convert torr to atm and mol to molecules
100 C
PV
113 þ 273 ¼ 386 K; PV ¼ nRT or n ¼
¼ 113 C;
180 F
RT
ð952 torrÞð1 atm=760 torrÞð3:29 L N2 Þ
¼ 0:130 mol N2
ð0:0821 L atm=mol KÞð386 KÞ
6:022 1023 molecule N2
¼ 7:83 1022 molecule N2
ð0:130 mol N2 Þ
1 mol N2
ð235 F 32 FÞ
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- Chapter 12 (c)
Temperature must be in K
0 þ 273 ¼ 273 K;
PV ¼ nRT or n ¼
PV
RT
ð0:624 atmÞð5:05 L Cl2 Þ
¼ 0:141 mol Cl2
ð0:0821 L atm=mol KÞð273 KÞ
6:022 1023 molecule Cl2
ð0:141 mol Cl2 Þ
¼ 8:49 1022 molecule Cl2
1 mol Cl2
The container with chlorine gas contains the largest number of molecules.
62. Assume 1 mol of each gas
(a)
SF6 ¼ 146:1 g=mol SF6
146:1 g
1 mol
d¼
¼ 6:52 g=L SF6
mol SF6
22:4 L
(b)
Assume 25 C and 1 atm pressure
298 K
Vðat 25 CÞ ¼ ð22:4 L C2 H6 Þ
¼ 24:5 L C2 H6
273 K
C2 H6 ¼ 30:07 g=mol
30:07 g
1 mol
¼ 1:23 g=L C2 H6
d¼
mol
24:5 L C2 H6
(c)
He at 80 C and 2.15 atm
ð1 mol HeÞð0:0821 L atm=mol KÞð193 KÞ
¼ 7:37 L He
2:15 atm
4:003 g
1 mol
d¼
¼ 0:543 g=L He
mol
7:37 L He
V¼
SF6 has the greatest density
63. (a)
(b)
Sample 2 has the higher density because it has more molecules in the same volume.
Sample 2 has the higher density because the molar mass of the individual molecules is larger.
64. (a)
Empirical formula. Assume 100. g starting material
80:0 g C
¼ 6:66 mol C
12:01 g=mol
20:0 g H
¼ 19:8 mol H
1:008 g=mol
6:66
¼ 1:00
6:66
19:8
¼ 2:97
6:66
Empirical formula ¼ CH3
Empirical mass ¼ 12:01 g þ 3:024 g ¼ 15:03 g
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- Chapter 12 (b)
Molecular formula:
2:01 g
1:5 L
22:4 L
¼ 30: g=molðmolar massÞ
mol
30: g=mol
¼ 2; Molecular formula is C2 H6
15:03 g=mol
(c)
Valence electrons ¼ 2ð4Þ þ 6 ¼ 14
H H
H C C H
H H
65. PV ¼ nRT
ð790 torrÞð1 atmÞ
(a)
ð2:0 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð298 KÞ
760 torr
n ¼ 0:085 molðtotal molesÞ
(b)
mol N2 ¼ total moles mol O2 mol CO2
¼ 0:085 mol 0:65 g O2
0:58 g CO2
32:00 g=mol 44:01 g=mol
mol N2 ¼ 0:085 mol 0:020 mol O2 0:013 mol CO2 ¼ 0:052 mol N2
28:02 g
ð0:052 mol N2 Þ
¼ 1:5 g N2
mol
(c)
0:020 mol O2
PO2 ¼ ð790 torrÞ
¼ 1:9 102 torr ðO2 Þ
0:085 mol
0:013 mol CO2
¼ 1:2 102 torr ðCO2 Þ
PCO2 ¼ ð790 torrÞ
0:085 mol
0:051 mol N2
PN2 ¼ ð790 torrÞ
¼ 4:7 102 torr ðN2 Þ
0:085 mol
! 2 CO2
66. 2 CO þ O2 Calculate the moles of O2 and CO to find the limiting reactant.
PV ¼ nRT
O2 : ð1:8 atmÞð0:500 L O2 Þ ¼ ðnÞð0:0821 L atm=mol KÞð288 KÞ
mol O2 ¼ 0:038 mol
800 mm Hg 1 atm
CO :
ð0:500 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð333 KÞ
760 mm Hg
mol CO ¼ 0:019 mol
Limiting reactant is CO
0:0095 mol O2 will react with 0:019 mol CO:
2 mol CO2
22:4 L
¼ 0:43 L CO2 ¼ 430 mL CO2
ð0:019 mol COÞ
mol
2 mol CO
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- Chapter 12 67. PV ¼ nRT or PV ¼
g
RT
molar mass
1:4 g 1000 cm3
¼ 1:4 103 g=L
cm3
L
1:4 103 g
9
ð0:0821 L atm=mol KÞðTÞ
1:3 10 atm ð1:0 LÞ ¼
2:0 g=mol
T¼
ð1:3 109 atmÞð1:0 LÞð2:0 g=molÞ
¼ 2:3 107 K
ð0:0821 L atm=mol KÞð1:4 103 gÞ
68. (a)
Assume atmospheric pressure of 14.7 lb=in.2 to begin with.
Total pressure in the ball ¼ 14:7 lb=in:2 þ 13 lb=in:2 ¼ 28 lb=in:2
PV ¼ nRT
1 atm
2
ð2:24 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð293 KÞ
ð28 lb=in: Þ
14:7 lb=in:2
n ¼ 0.18 mol air
(b)
(c)
mass of air in the ball
molar mass of air is about 29 g/mol
29 g
m ¼ ð0:18 molÞ
¼ 5:2 g air
mol
Actually the pressure changes when the temperature changes. Since pressure is directly
proportional to moles we can calculate the change in moles required to keep the pressure the
same at 30 C as it was at 20 C.
PV ¼ nRT
1 atm
2
ð28 lb=in: Þ
ð2:24 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð303 KÞ
14:7 lb=in:2
n ¼ 0.17 mol of air required to keep the pressure the same at 30 C.
0.01 mol air (0.18 – 0.17) must be allowed to escape from the ball.
29 g
ð0:01 mol airÞ
¼ 0:29 g or 0:3 g air must be allowed to escape.
mol
69. Use the combined gas laws to calculate the bursting temperature (T2).
P1 V1 P2 V2
¼
T1
T2
T2 ¼
P1 ¼ 65 cm
76 cm
P2 ¼ 1:00 atm
¼ 76 cm
1 atm
V1 ¼ 1:75 L
T1 ¼ 20 Cð293 KÞ
V2 ¼ 2:00 L
T2 ¼ T2
P2 V2 T1 ð76 cmÞð2:00 LÞð293 KÞ
¼ 392 Kð119 CÞ
¼
ð65 cmÞð1:75 LÞ
P1 V1
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- Chapter 12 70. To double the volume of a gas, at constant pressure, the temperature (K) must be doubled.
V1 V2
¼
T1 T2
V1 2 V1
¼
T1
T2
V2 ¼ 2 V1
T2 ¼
2 V1 T 1
V1
T2 ¼ 2 T1
T2 ¼ 2ð300: KÞ ¼ 600: K ¼ 327 C
71. V ¼ volume at 22 C and 740 torr
2 V ¼ volume after change in temperatureðP constantÞ
V ¼ volume after change in pressureðT constantÞ
P 1 V1
Since temperature is constant, P1 V1 ¼ P2 V2 or P2 ¼
V2
2V
P2 ¼ ð740 torrÞ
¼ 1:5 103 torr ðpressure to change 2 V to VÞ
V
72. Use the combined gas laws
ðP1 ÞðV1 Þ ðP2 ÞðV2 Þ
ðT1 ÞðP2 ÞðV2 Þ
¼
or T2 ¼
ðT1 Þ
ðT2 Þ
ðP1 ÞðV1 Þ
Since the volume stays constant, V1 = V2 and the equation reduces to T2 ¼
T2 ¼
ðT1 ÞðP2 Þ
ðP1 Þ
ð500: torrÞð295 KÞ
¼ 211 K ¼ 62 C
700: torr
73. Use the combined gas laws
ðP1 ÞðV1 Þ ðP2 ÞðV2 Þ
¼
ðT1 Þ
ðT2 Þ
or T2 ¼
ðT1 ÞðP2 ÞðV2 Þ
ðP1 ÞðV1 Þ
The volume of the tires remains constant (until they burst) so V1 = V2 and the equation reduces to
ðT1 ÞðP2 Þ
T2 ¼
ðP1 Þ
71:0 F ¼ 21:7 C ¼ 295 K
T2 ¼
ð44 psiÞð295 KÞ
¼ 433 K ¼ 160 C ¼ 320 F
30: psi
74. Use the combined gas laws.
P1 V1 P2 V2
¼
T1
T2
P2 ¼
or P2 ¼
P1 V1 T2
V2 T 1
P1 and T1 are at STP
ð1:00 atmÞð800: mLÞð303 KÞ
¼ 3:55 atm
ð250: mLÞð273 KÞ
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- Chapter 12 75. 1 mol of a gas occupies 22.4 L at STP
1 mol
22:4 SO2
¼ 3:20 L SO2
ð9:14 g SO2 Þ
64:07 g 1 mol SO2
76. Use the combined gas law
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
P1 V1 T2
P2 T1
First calculate the volume at STP.
V2 ¼
ð400: torrÞð600: mL N2 OÞð273 KÞ
¼ 275 mL N2 O ¼ 0:275 L N2 O
ð760: torrÞð313 KÞ
At STP, a mole of any gas has a volume of 22.4 L
1 mol
6:022 1023 molecules
ð0:275 L N2 OÞ
¼ 7:39 1021 molecules N2 O
22:4 L
1 mol
Each molecule of N2O contains 3 atoms, so:
3 atoms
21
7:39 10 molecules N2 O
¼ 2:22 1022 atoms
1 molecule N2 O
77. Use the combined gas laws
ðP1 ÞðV1 Þ ðP2 ÞðV2 Þ
¼
ðT1 Þ
ðT2 Þ
or
P2 ¼
ðP1 ÞðT2 ÞðV1 Þ
ðT1 ÞðV2 Þ
Since the volume stays constant (unless it does burst), V1 = V2 and the equation reduces to
ðP1 ÞðT2 Þ
P2 ¼
ðT1 Þ
T1 ¼ 25 C þ 273 ¼ 298 K
T2 ¼ 212 F ¼ 100 C ¼ 373 K
P2 ¼
ð32 lb=in:2 Þð373 KÞ
¼ 40: lb=in:2
298 K
At 212 F the tire pressure is 40. lb/in.2
The tire will not burst.
78. A column of mercury at 1 atm pressure is 760 mm Hg high. The density of mercury is 13.6 times that of
water, so a column of water at 1 atm pressure should be 13.6 times as high as that for mercury.
ð760 mmÞð13:6Þ ¼ 1:03 104 mmð33:8 ftÞ
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- Chapter 12 79. Use the ideal gas equation
PV ¼ nRT
n¼
RT
PV
Change 2:20 103 lb=in:2 to atmosphere
1 atm
3
2
¼ 150: atm
2:20 10 lb=in:
14:7 lb=in:2
n¼
ð150: atmÞð55 L O2 Þ
¼ 3:3 102 mol O2
0:0821 L atm
ð300: KÞ
mol K
80. The conversion is: m3 ! cm3 ! mL ! L ! mol
3 100 cm
1 mL
1L
1 mol
3
ð1:00 m Cl2 Þ
¼ 44:6 mol Cl2
1m
1 cm3
1000 mL 22:4 L
81. First calculate the moles of gas and then convert moles to molar mass.
1 mol
¼ 0:0250 mol
ð0:560 LÞ
22:4 L
1:08 g
¼ 43:2 g=molðmolar massÞ
0:0250 mol
82. The conversion is: g=L ! g=mol
1:78 g 22:4 L
¼ 39:9 g=molðmolar massÞ
L
mol
83. PV ¼ nRT
nRT ð0:510 mol H2 Þð0:0821 L atm=mol KÞð320: KÞ
(a) V ¼
¼
¼ 8:4 L H2
P
1:6 atm
(b)
(c)
n¼
PV ð600 torrÞð1 atm=760 torrÞð16:0 L CH4 Þ
¼
¼ 0:513 mol CH4
RT
ð0:0821 L atm=mol KÞð300: KÞ
The molar mass for CH4 is 16.04 g=mol
ð16:04 g=molÞð0:513 molÞ ¼ 8:23 g CH4
g
PV ¼ nRT, but n ¼ where M is the molar mass and g is the grams of the gas.
M
gRT
. To determine density, d ¼ g=V.
Thus, PV ¼
M
gRT
g
g PM
Solving PV ¼
for produces ¼
.
M
V
V RT
g
ð4:00 atmÞð44:01 g=mol CO2 Þ
d¼ ¼
¼ 8:48 g=L CO2
V ð0:0821 L atm=mol KÞð253 KÞ
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- Chapter 12 -
(d)
Since d ¼
M¼
g PM
¼
from part (c), solve for M (molar mass)
V RT
dRT ð2:58 g=LÞð0:0821 L atm=mol KÞð300: KÞ
¼
¼ 63:5 g=mol ðmolar massÞ
P
1:00 atm
! C2 H4 F2 ðgÞ
84. C2 H2 ðgÞ þ 2 HFðgÞ 1:0 mol C2 H2 ! 1:0 mol C2 H4 F2
1 mol C2 H4 F2
ð5:0 mol HFÞ
¼ 2:5 mol C2 H4 F2
2 mol HF
C2H2 is the limiting reactant. 1.0 mol C2H4F2 forms, no moles C2H2 remain.
According to the equation, 2.0 mol HF yields 1.0 mol C2H4F2. Therefore,
5:0 mol HF 2:0 mol HF ¼ 3:0 mol HF unreacted
The flask contains 1.0 mol C2H4F2 and 3.0 mol HF when the reaction is complete.
The flask contains 4.0 mol of gas.
nRT ð4:0 molÞð0:0821 L atm=mol KÞð273 KÞ
¼
¼ 9:0 atm
V
10:0 L
3 mol H2
22:4 L
85. ð8:30 mol AlÞ
¼ 279 L H2 at STP
mol
2 mol Al
P¼
86. Assume 100. g of material to start with. Calculate the empirical formula.
1 mol
7:14
C
ð85:7 gÞ
¼ 1:00 mol
¼ 7:14 mol
12:01 g
7:14
1 mol
14:2
H
ð14:3 gÞ
¼ 14:2 mol
¼ 1:99 mol
1:008 g
7:14
The empirical formula is CH2. To determine the molecular formula, the molar mass
must be known.
2:50 g 22:4 L
¼ 56:0 g=molðmolar massÞ
L
mol
56:0
¼4
The empirical formula mass is 14:0
14:0
Therefore, the molecular formula is ðCH2 Þ4 ¼ C4 H8
87. 2 COðgÞ þ O2 ðgÞ ! 2 CO2 ðgÞ Determine the limiting reactant
2 mol CO2
ð10:0 mol COÞ
¼ 10:0 mol CO2 ðfrom COÞ
2 mol CO
2 mol CO2
ð8:0 mol O2 Þ
¼ 16 mol CO2 ðfrom O2 Þ
1 mol O2
CO : the limiting reactant,
O2 : in excess, 3:0 mol O2 unreacted:
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- Chapter 12 (a)
10.0 mol CO react with 5.0 mol O2
10.0 mol CO2 and 3.0 mol O2 are present, no CO will be present.
(b)
P¼
nRT ð13 molÞð0:0821 L atm=mol KÞð273 KÞ
¼
¼ 29 atm
V
10: L
D
! 2 KC1ðsÞ þ 3 O2 ðgÞ
88. 2 KClO3 ðsÞ First calculate the moles of O2 produced. Then calculate the grams of KClO3 required
to produce the O2. Then calculate the % KC1O3.
1 mol
ð0:25 L O2 Þ
¼ 0:011 mol CO2
22:4 L
2 mol KClO3
122:6 g
ð0:011 mol O2 Þ
¼ 0:90 g KClO3 in the sample
mol
3 mol O2
0:90 g
ð100Þ ¼ 75% KClO3 in the mixture
1:20 g
89. Assume 1.00 L of air. The mass of 1.00 L of air is 1.29 g.
P 1 V1 P 2 V2
¼
T1
T2
V2 ¼
d¼
P1 V1 T2 ð760 torrÞð1:00 LÞð290 KÞ
¼
¼ 1:8 L
ð450 torrÞð273 KÞ
P2 T1
m 1:29 g
¼
¼ 0:72 g=L
V
1:8 L
90. Each gas behaves as though it were alone in a 4.0 L system.
(a) After expansion: P1 V1 ¼ P2 V2
For CO2
For H2
(b)
P1 V1 ð150: torrÞð3:0 LÞ
¼
¼ 1:1 102 torr
4:0 L
V2
P1 V1 ð50: torrÞð1:0 LÞ
¼ 13 torr
P2 ¼
¼
4:0 L
V2
P2 ¼
Ptotal ¼ PH2 þ PCO2 ¼ 110 torr þ 13 torr ¼ 120 torr ð2 sig: figuresÞ
91. Use the combined gas laws
ðP1 ÞðV1 Þ ðP2 ÞðV2 Þ
¼
ðT1 Þ
ðT2 Þ
P1 ¼ 40:0 atm
V1 ¼ 50:0 L
T1 ¼ 25 C ¼ 298 K
or P2 ¼
ðP1 ÞðT2 ÞðV1 Þ
ðT1 ÞðV2 Þ
P2 ¼ unknown
V2 ¼ 50:0 L
T2 ¼ 25 C þ 152 C ¼ 177 C ¼ 450 K
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- Chapter 12 -
P2 ¼
ð40:0 atmÞð450 KÞð50:0 LÞ
¼ 60:4 atm
ð298 KÞð50:0 LÞ
(Note that the volume canceled out of the expression because it stays constant.)
92. You can identify the gas by determining its density.
mass of gas ¼ 1:700 g 0:500 g ¼ 1:200 g
volume of gas: Charles law problem. Correct volume to 273 K
V1 V2
¼
T1 T2
V2 ¼
V1 T2 ð0:4478 LÞð273 KÞ
¼ 0:3785 L
¼
323 K
T1
m
1:200 g
¼
¼ 3:170 g=L
V 0:3785 L
gas is chlorine (see Table 12.4)
d¼
93. 1st step: find the volume of CO2 that must be produced
55:0 L CO2
volume CO2 ¼ ðvolume of batterÞð55:0%Þ ¼ ð1:32 L batterÞ
¼ 0:726 L CO2
100 L batter
2nd step: find the moles of CO2 needed to produce the necessary rise in the cupcakes
PV
or n ¼
RT
1 atm
P ¼ ð738 torrÞ
¼ 0:971 atm
760 torr
V ¼ 0:726 L
T ¼ 325 F ¼ 163 C ¼ 436 K
PV ¼ nRT
n
¼
ð0:971 atmÞð0:726 LÞ
¼ 0:0197 mol CO2
ð0:0821 L atm=mol KÞð436 KÞ
3rd step: Calculate the mass of sodium bicarbonate which will produce the CO2 needed
3 NaHCO3 þ H3 C6 H5 O7 ! Na3 C6 H5 O7 þ 3 H2 O þ 3 CO2
ð0:0197 mol CO2 Þ
3 mol NaHCO3
84:02 g
¼ 1:66 g NaHCO3
1 mol
3 mol CO2
4th step: calculate mass of sodium bicarbonate needed to add since only 63.7% will decompose to form
carbon dioxide
100 g NaHCO3 added
ð1:66 g NaHCO3 reactedÞ
¼ 2:61 g NaHCO3 added
63:7 g NaHCO3 reacted
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