Chemistry 101
ANSWER KEY
REVIEW QUESTIONS
Chapter 5
1. Determine the pressure of the gas (in mmHg) in the diagram
below, given atmospheric pressure= 0.975 atm.
760 mmHg
= 741 mmHg
1 atm
10 mmHg
h = 52 cmHg x
= 520 mmHg
1 cmHg
Pgas =Patm - h = 741 mmHg - 520 mmHg = 221 mmHg
Patm = 0.975 atm x
2. A sample of oxygen gas has a volume of 26.7 L at 752 mmHg and 20C. What is the
volume of this gas at 1.30 atm and 20C?
V2 = V1 x
P1
752 mmHg
= 26.7 L x
= 20.3 L
760 mmHg
P2
1.30 atm x
1 atm
3. A 35.8 L cylinder of Argon gas is connected to and transferred into an evacuated 1875-L
tank at constant temperature. If the final pressure in the tank is 721 mmHg, what must
have been the original pressure (in atm) in the cylinder?
P1 = P2 x
V2
1 atm
1875 L
= (721 mmHg x
)x
= 49.7 atm
V1
760 mmHg
35.8 L
4. A 34.0-L cylinder contains 305 g of oxygen gas at 22C. How many grams of gas must
be released to reduce the pressure in the cylinder to 1.15 atm if the temperature remains
constant?
Mass of gas present at 1.15 atm
PV
(1.15 atm )(34.0 L )
=
= 1.614 mol
RT (0.0821 L atm )(295 K )
mol K
32.0 g
mass of oxygen = 1.614 mol x
= 51.6 g
1 mol
Mass of gas that must be removed
n=
305 g – 51.6 g = 253 g
1
5. At STP, 0.280 L of a gas weighs 0.400 g. Calculate the molar mass of this gas.
moles of gas = 0.280 L x
molar mass =
1 mol
= 0.0125 mol
22.4 L
0.400 g
= 32.0 g/mol
0.0125 mol
6. Calculate the density of HBr gas in g/L at 733 mmHg and 46C.
733
atm)(80.91 g/mol)
PM
760
d=
=
) = 2.98 g/L
RT (0.0821Latm/molK)(319 K
(
7. A mixture of 4.00 g of hydrogen and 10.0 g of helium are in a 4.30-L flask at 0C. What
is the total pressure of the container and the partial pressures of each gas?
1 mol
= 1.98 mol
2.02 g
1 mol
mol He = 10.0 g x
= 2.50 mol
4.00 g
Total mol = 1.98 + 2.50 = 4.48 mol
mol H 2 = 4.00 g x
L atm
)(273 K )
nRT
mol
K
PTotal =
=
= 23.4 atm
V
4.30 L
1.98 mol
2.50 mol
XH2 =
=0.442
X He =
=0.558
4.48 mol
4.48 mol
PH2 = X H2 PTot = (0.442)(23.4 atm) = 10.3 atm
(4.48 mol )(0.0821
PHe = X He PTot = (0.558)(23.4 atm) = 13.1 atm
2
8. Life rafts and weather balloons can be inflated by the reaction shown below:
CaH2 (s) + 2 H2O (l)  Ca(OH)2 (aq) + 2 H2 (g)
How many grams of CaH2 are needed to produce 10.0 L of hydrogen gas at 740
mmHg and 23C?
1 atm
(740 mmHg x
)(10.0 L )
PV
760 mmHg
n H2 =
=
= 0.4007 mol
L atm
RT
(0.0821
)(296 K )
mol K
1 mol CaH 2 42.10 g
0.4007 mol H 2 x
x
= 8.43 g CaH 2
2 mol H 2
1 mol
9. Nitroglycerin, an explosive compound, decomposes according to the equation below:
4 C3H5(NO3)3 (s)  12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g)
Calculate the total volume of gases produced at 1.2 atm and 26C when 260 g of
nitroglycerine is decomposed.
1 mol
29 mol gas
x
= 8.30 mol gas
227.1 g
4 mol NG
L atm
(8.30 mol )(0.0821
)(299 K )
nRT
mol K
V=
=
= 170 L
P
1.2 atm
260 g NG x
10. A 1.65-g sample of Al is reacted with excess HCl and the hydrogen produced is
collected over water at 25C at a barometric pressure of 744 mmHg. What volume of
hydrogen gas is produced in this reaction? (Vapor pressure of water at 25C is 23.8
mmHg)
2 Al (s) + 6 HCl (aq)  2 AlCl3 (aq) + 3 H2 (g)
PH2 = Patm - PH 2O = 744 mmHg - 23.8 mmHg = 720.2 mmHg
3 mol H 2
1 mol
x
= 0.09173 mol H 2
27.0 g 2 mol Al
L atm
(0.09173 mol )(0.0821
)(298 K )
nRT
mol
K
V=
=
= 2.37 L
1 atm
P
720.2 mmHg x
760 mmHg
1.65 g Al x
3
11. Nitric oxide reacts with oxygen gas as shown below:
2 NO (g) + O2 (g)  2 NO2 (g)
Initially NO and O2 are separated as shown in the diagram below. When the valve is opened
the reaction quickly goes to completion. Determine the identity of the gases that remain at
the end of the reaction and their partial pressure. Assume temperature remains at 25C.
Since the system is at constant temperature, the moles of gas is proportional to the
product of pressure and volume as shown below:
PV = n RT
PV  n
at constant temp.
Before reaction
mol NO α (4.00 L)(0.500 atm) = 2.00
mol O 2 α (2.00 L)(1.00 atm) = 2.00
2 NO + O 2
2 NO 2
Initial
React
2.00
2.00
2.00
1.00
0
2.00
End
0
1.00
2.00
At end of reaction, total volume is 6.00 L. Therefore,
mol 1.00
=
= 0.167 atm
V
6.00
mol 2.00
α
=
= 0.333 atm
V
6.00
PO2 α
PNO2
4
12. A 4.85-g sample of solid ammonium chloride is placed in a 1.50-L evacuated flask and
heated until it decomposes, as shown below:
NH4Cl (s)  NH3 (g) + HCl (g)
After the reaction has completed, the total pressure in the flask is measured as 4.40 atm at
202C. Based on this information, what percent of the ammonium chloride decomposed?
Ptotal = PNH3 + PHCl
and
PNH3 = PHCl
4.40 atm
= 2.20 atm
2
PV
(2.20 atm )(1.50 L )
n NH3 =
=
= 0.0846 mol
RT (0.0821 L atm )(475 K )
mol K
1 mol NH 4Cl 53.50 g
0.0846 mol NH 3 x
x
= 4.526 g NH 4Cl reacted
1 mol NH 3
1 mol
4.526 g
% decomposed =
x100 = 93.3%
4.85 g
PNH3 =
13. A mixture of gases containing 12.45 g of H2, 60.67 g of N2 and 2.380 g of NH3 are
placed in a 10.00 L container at 90°C. What is the total pressure (in atm) and partial
pressure of each component in the gas mixture?
1 mol
= 6.176 mol
2.016 g
1 mol
mol N 2 = 60.67 g x
= 2.165 mol
28.02 g
1 mol
mol NH 3 = 2.380 g x
= 0.1397 mol
17.04 g
Total mol = 6.176 + 2.165 + 0.1397 = 8.481 mol
mol H 2 = 12.45 g x
L atm
(8.481 mol )(0.0821
)(363 K )
nRT
mol
K
PTotal =
=
= 25.28 atm
V
10.00 L
6.176 mol
X H2 =
=0.7282
PH2 = X H2 PTot = (0.7282)(25.28 atm) = 18.41 atm
8.481 mol
2.165 mol
X He =
=0.2553
PHe = X He PTot = (0.2553)(25.28 atm) = 6.454 atm
8.481 mol
0.1397 mol
X NH3 =
=0.01647
PNH3 = X NH3PTot = (0.01647)(25.28 atm) = 0.4164 atm
8.481 mol
5
14. Which has a higher average speed, H2 at 150 K or He at 375C?
u H2 =
3RT
=
M
3 (8.314 J/molK)(150 K)
= 1360 m/s
2.02x10-3 kg/mol
u He =
3RT
=
M
3 (8.314 J/molK)(648 K)
= 2010 m/s
4.00x10-3 kg/mol
He at 375 oC has higher average speed
15. A big-league fastball travels at about 45.0 m/s. At what temperature (C) do
helium atoms have this same average speed?
u He =
3RT
=
M
3 (8.314 J/molK)(T)
= 45.0 m/s
4.00x10-3 kg/mol
squaring both sides,
3 (8.314 J/molK)(T)
= 2025 m 2 /s 2
-3
4.00x10 kg/mol
T = 0.325 K
T = 0.325 -273.15 = -272.83 oC
16. The surface temperature of Venus is about 1050 K, and the pressure is about 75.0 earth
atmospheres. Assuming that these conditions represent a Venusian “STP”, what is the
“standard” molar volume of a gas on Venus?
V=
nRT
=
P
L atm
)(1050 K )
mol K
= 1.15 L
75.0 atm
(1.00 mol )(0.0821
6
17. Gaseous iodine IF5 can be prepared by the reaction of solid iodine and gaseous fluorine:
I2 (s) + 5 F2 (g)  2 IF5 (g)
A 5.00-L flask is charged with 10.0 g of I2 and 10.0 g of F2, and the reaction proceeds
until one of the reagents is completely consumed. After the reaction is complete the
temperature in the flask is 125C. What is the partial pressure of IF5 in the flask?
Before reaction,
1 mol
= 0.0394 mol
253.8 g
1 mol
mol F2 = 10.0 g x
= 0.263 mol
38.00 g
mol I 2 = 10.0 g x
I 2 + 5 F2
2 IF5
Initial
0.0394
0.263
0
React
0.0394
0.197
0.0788
0.066
0.0788
End
0
At end of reaction,
Total moles = 0.066 + 0.0788 = 0.1448 mol
n RT
(0.1448 mol)(0.821 Latm/molK)(398 K)
Ptotal = total
=
= 0.946 atm
V
5.00 L
0.0788 mol
XIF5 =
=0.544
PIF5 = X IF5 Ptotal = (0.544)(0.946 atm) = 0.515 atm
0.1448 mol
7