1
Chemistry 120 . (Chapter 9)
1.
Carry the following conversions between pressure units.
a)
0.936 atm to Torr
b)
200 mmHg to atm
a) Torr = (0.936 atm) (760 Torr) = 711 Torr
1 atm
c)
1.25 bar to atm
b) atm = (200 mmHg) (1atm) = 0.263 atm
760 mmHg
c)
atm = (1.25 bar) (1 atm)
1.013 bar
2.
A sample of xenon is held at constant temperature and occupies 882 mL at 752 Torr.
Determine the volume of xenon at a) 719 Torr
b) 1.38 atm
a) P1V1 = P2V2
P1V1
P2
= 1.23 atm
= V2
(752 Torr) (882mL)
719 Torr
= 922 mL
Note: There is no need to change the units of pressure to units of atm. Only temperature must
be change to absolute temperature scale (Kelvin) because in a ratio of temperatures a value of
0oC can give a result of zero.
b) P1 (atm) = (752 Torr) (1atm) = 0.9895 atm
760 Torr
P1V1 = P2V2
3.
P1V1
P2
= V2
(0.9895 atm) (882mL)
1.38 atm
= 632 mL
A decompression chamber used by deep-sea divers has a volume of 10m3 and operates at
an internal pressure of 4.50 atm. What volume in cubic meters, would the air chamber
occupy if it were at normal atmospheric pressure (1 atm), assuming no temperature
change?
V1 = 10 m3
P1 = 4.50 atm
P2 = 1 atm
V2 = ?
P1V1 = P2V2
P1V1
P2
= V2
(4.50 atm)(10 m3) = 45 m3
1 atm
2
4.
Oxygen used in respiration therapy is stored at room temperature (25oC) under a pressure
of 1.50 x 102 atm in a gas cylinder with a volume of 43.0 L.
a)
What volume would the gas occupy at pressure of 750.0 torr? Assume no temperature
change.
V1 = 43 L
P1 = 150 atm
P2 = 750 Torr
V2 = ?
P1V1 = P2V2
P1V1
P2
= V2
750 Torr (1 atm) = 0.987 atm
760 Torr
(150 atm)(43 L) = 6534 L ~ 6.5x 103L
0.987 atm
b)
If the oxygen flow is adjusted to 8.00 L per minute at room temperature and 750.0 Torr,
how long will the tank of gas last?
Time for tank to last = (6534) (1 min) = 817 min
8L
5.
A gas at a temperature of 99.8 oC occupies a volume of 641 mL. What will the volume be
at a temperature of 5.0 oC, assuming no change in pressure?
T1 = 99.8oC + 273 K = 372.8 K
V1 = 0.641 L
T2 = 278 K
V2 = ?
V1T2 = V2T1
V1T2
T1
(0.641 L)(278 K)
372.8K
6.
= V2
= 478 mL
What is the mass in kilograms of 4.55 x 103 L of methane (CH4) gas at STP?
P = 1 atm
V = 4.55 x 103 L
T = 273 K
n=?
R = 0.0821 L atm
mol● K
PV = nRT
PV
= n
RT
= 203 mole
(1atm)(4.55 x 103L)
(0.0821 L atm)(273K)
mol● K
mol CH4 = 203 moles
grams CH4 = (203 mole CH4) (16 grams) = 3248 grams ~ 3,25 kg CH4
1 mole
3
7.
What is volume in, milliliters, of 88.3 mg of carbon dioxide gas at STP?
mole CO2 = (88.3 mg)(1 gram) ( 1 mole) = 0.00201 mole
(1000 mg) (44 grams)
8.
At STP,
P = 1atm and T = 273K
PV = nRT
V = nRT
V = (0.00201 mole) (0.0821 L atm) (273K) = 45.1mL
mol● K
1atm
How many grams of gas must be released from a 45.2L sample of N2 gas at STP to reduce
the volume to 35.0 L at STP?
V = 45.2 L - 35.0L = 10.2 L
P = 1 atm
T= 273 K
n =?
PV = nRT
PV
Rv
RT
= n
(1 atm)(10.2L)
= 0.455 moles
0.0821 L atm (273 K)
mol● K
Grams of N2 = (0.455 mole N2) (28 grams) = 12.7 grams
1 mole N2
9.
At 25oC, the pressure in a gas cylinder containing O2 gas is 5.05 atm. To maintain constant
pressure of 5.05 atm, how many moles of O2 gas should be released when the temperature is
raised to 235oC?
Note this question is incomplete – (the cylinder should initially have 5.00 mole O2 gas).
P1 = P2 = 5.05 atm; V1 = V2; T1 = 298K ; T2 = 508K
PV = nRT (pressure and volume are constant)\
n1T1 = n2T2
n1T1
T2
= n2
(5.00mol)(298K)
508K
= 2.93 mole
Mole of O2 that needs to be released = 5.00 mole - 2.93 mole = 2.07 mole O2
4
10.
Calculate the molecular mass of a liquid that, when vapourized at 98oC and 756 Torr, gives
139 mL of vapour with a mass of 0.808 grams.
V = 0.139 L
PV = nRT
P = (756 Torr) x (1 atm) = 0.9947 atm
760 Torr
T= 371 K
n =?
(0.9947 atm)(0.139L)
0.0821 L atm (371 K)
mol● K
Molecular mass =
11.
PV
RT
= n
= 0.00454 mole
(0.808 grams) = 178 grams/mol
0.00454 mole
A liquid hydrocarbon is found to be 8.75% H by mass. A 1.261 gram vapourized sample
of the hydrocarbon has a volume of 435 mL at 115oC and 761 mmHg. What is the
molecular formula of this hydrocarbon?
V = 0.435 L
PV = nRT
P = (761 mm Hg) x (1 atm) = 1.0013 atm
760 mmHg
T= 388 K
n =?
(1.0013 atm)(0.435L)
0.0821 L atm (388 K)
mol● K
Molecular mass =
PV
RT
= 0.01367 mole
(1.261 grams) = 92.2 grams/mol
0.01367 mole
There is 8.75 % H by mass.
So, Number of hydrogens = (92 g/mole) x (0.0875) = 8 Hydrogens
The rest is made of carbons:
C7H8 is molecular formula
Hydrocarbons only have carbons and hydrogens
= n
5
12.
The combustion of Butane (C4H10) releases CO2 (g) and H2O (l).
a) Write a balanced chemical equation for the combustion of butane.
C4H10(g)_ + 13/2 O2 (g) →
4CO2 (g) + 5H2O (g)
b) If 10.0 mL of butane was allowed to react until completion at 25oC and 1 atm, how
many grams of CO2 is produced?
V = 0.010 L
P = 1 atm
T= 25oC = 298 K
n =?
PV = nRT
PV
RT
= n
(1 atm)(0.010L)
= 4.087 x 10-4 mole butane
0.0821 L atm (298 K)
mol● K
-4
Mole CO2 produced = (4.087 x 10 mole butane)(4 mole CO2) = 1.635 x 10-3
1 mole butane
Grams of CO2 Produced = (1.635 x 10-3 mole CO2)(44grams)
mole CO2
= 0.072 grams
c) On a different day, the combustion reaction was performed and CO2 was
collected over water at 28oC. The wet gas occupies a volume of 4.5 L at a total
pressure of 748 mmHg. How many mole of CO2 was collected and how many grams
of Butane is theoretically required?
Ptotal = 748 mmHg PH2O = 28.35 mmHg (page 604)
Ptotal = PH2O + PCO2
PCO2 = Ptotal – PH2O
= 748 mmHg – 28.35 mmHg = 719.65 mmHg
PV
= n
V = 4.5 L
PV = nRT
RT
P = (719.65mm Hg) x (1 atm) = 0.947 atm
760 mmHg
T= 28oC + 273 = 301 K
n =?
(0.947 atm)(4.5L)
= 0.1724 mole CO2
0.0821 L atm (301 K)
mol● K
Mole butane used = (0.1724 mole CO2)(1 mole butane) = 0.0431 mole butane
4 mole CO2
Grams of butane used =
(0.0431 mole butane)(58 grams)
mole butane
= 2.5 grams
6
13.
One mole of any gas has a volume of 22.414 L at STP. What are the densities (grams per
liter) at STP?
a) CH4
b) CO2
c) UF6
molar mass of CH4 = 16 g
molar mass of CO2 = 44 g
molar mass of UF6 = 352 g
1 mole CH4
1 mole CO2
1 mole UF6
Density of CH4
= (16 g )
1 mole CH4
x ( 1 mole CH4) = 0.71384 g/L
22.414 L
molar mass of CO2 = (44 g)
x (1 mole CO2)
1 mole CO2
22.414L
molar mass of UF6 = (352 g)
1 mole UF6
14.
x ( 1 mole UF6)
22.414 L
= 1.9631 g/L
= 15.704 g/L
Nitrogen oxide is a pollutant commonly found in smokestack emissions. One way to
remove NO, is to react it with ammonia.
4NH3 (g) + 6NO(g) → 5N2 (g) + 6H2O (l)
How many liters of ammonia are required to change 12.8L of nitrogen oxide to nitrogen
gas? Assume 100% yield and that all gases are measured at the same temperature and
pressure.
V ∝ n (Temperature and Pressure are constant)
Volume NH3 = (12.8 L NO gas) (4 NH3)
(6 NO)
= 8.53 L
7
15.
When acetylene (C2H2) is burned in oxygen, Carbon dioxide and Steam are produced. A
sample of acetylene with a volume of 7.50L and a pressure of 1.00 atm is burned in excess
oxygen at 225oC. The products are transferred without loss to a 10.0L flask at the same
temperature.
a) Write a balanced chemical equation for the reaction
C2H2(g)_ + 5/2 O2 (g) →
2CO2 (g) + H2O (g)
b) What is the total pressure of the products in the 10.0L flask (H2O(g) - steam is gas)
PV
RT
PV = nRT
V = 7.50 L
P = 1 atm
T= 225oC + 273 = 498 K
n =?
= n
(1 atm)(7.50 L)
= 0.1834 mole acetylene
0.0821 L atm (498 K)
mol● K
Mole CO2 produced = (0.1834 mole C2H2)(2 mole CO2) = 0.3668 mole CO2
1 mole C2H2
Mole H2O produced = (0.1834 mole C2H2)(2 mole H2O) = 0.1834 mole H2O
1 mole C2H2
Total moles of gas produced. = 0.3688 + 0.1834 = 0.5522 mole (complete combustion)
V = 10.0 L
P= ?
T= 225oC + 273 = 498 K
n = 0.5522 mole
Ptotal =
Total pressure :
P = nRT
V
(0.5522 mole ) ( 0.0821 L atm )(498 K) = 2.258 atm
mol● K
10.0L
c) What is the partial pressure of each or the products in the flask?
PCO2 = χCO2 Ptotal
χCO2 =
0.3688 mole CO2
0.3688 mole CO2 + 0.1834 mole H2O
= 0.6679
χH2O = 1 – 0.6679 = 0.3321
χ = mole fraction
χ CO2 = mole CO2
mole CO2 + mole H2O
PCO2 = (0.6679)(2.258 atm) = 1.508 atm
PH2 O = (0.3321)(2.258 atm) = 0.750 atm
8
16.
Nitrogen can react with steam to form ammonia and Nitrogen oxide (NO) gases. A 20.0L
sample of nitrogen at 173oC and 772 mmHg is made to react with excess steam. The
products are collected at room temperature (25oC) in an evacuated flask with a volume of
15.0L.
a) Write a balanced equation for the reaction.
5 N2(g)_ + 6H2O (g)
→
4 NH3 (g) + 6NO (g)
b) What is the total pressure of the products in the collecting flask after the reaction is
complete?
V = 20.0 L
P = (772 mmHg) x (1 atm) 1.016 atm
760 mmHg
PV
RT
PV = nRT
= n
T= 173oC + 273 = 446 K
mole N2 n =?
(1.016 atm)(20.0 L)
= 0.5549 mole N2
0.0821 L atm (446 K)
mol● K
Mole NH3 produced = (0.5549 mole N2)(4 mole NH3) = 0.4439 mole NH3
5 mole C2H2
Mole NO produced = (0.5549 mole N2)(6 mole NO) = 0.6659 mole NO
5 mole N2
Total moles of gas produced. = 0.4439 + 0.6659 = 1.1098 mole (complete reaction)
V = 15.0 L
P= ?
T= 25oC + 273 = 298 K
n = 1.1098 mole
Ptotal =
Total pressure :
P = nRT
V
(1.1098 mole ) ( 0.0821 L atm )(298 K) = 1.81 atm
mol● K
15.0L
c) What is the partial pressure of each of the products in the flask?
PNH3 = χ NH3 Ptotal
χCO2 =
0.4439 mole NH3
0.4439 mole NH3 + 0.6659 mole NO
= 0.400
χH2O = 1 – 0.400 = 0.600
χ = mole fraction
χ NH3 = mole NH3
mole NH3 + mole NO
PNH3 = (0.400)(1.81atm) = 0.724 atm
PN O = (0.600)(1.81 atm) = 1.086 atm