CLAS – Ch 5 - Key
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1. The valve between two tanks is opened. See below. Calculate the ratio of partial pressures (O2:Ne) in the
container after the valve is opened. a. 1.31 b. 1.60 c. 2.10 d. 0.477 e. 0.615
Each gas is affected by the valve opening⇒
P1V1
n1T1
= Pn VT
2
2
2 2
where n and T are constant
P1V1
2
2
V2
(8.64 atm)(3.25 L)
2
(3.25L+2.48L)
solving for P ⇒ P =
for O2 ⇒ P =
P2 = 4.9 atm
atm)(2.48 L)
for Ne ⇒ P2 = (5.4
= 2.3 atm
(3.25L+2.48L)
atm
The ratio of partial pressures ⇒ 𝑃𝑃𝑂2 = 4.9
= 2.1
2.3 atm
𝑁𝑒
2. In an experiment 300 m3 of methane is collected over water at 785 torr and 65 °C. What is the volume of the dry gas
(in m3) at STP? The vapor pressure of water at 65 °C is 188 torr..
When a gas is collected over water there will be water vapor in the collection chamber. A dry gas implies that the
water vapor has been removed.
The PCH4 = Ptotal – PH2O = 785 torr – 188 torr = 597 torr
At STP the temperature and pressure are 273K and 760 torr respectively.
PV
Using the combined gas law where n is constant Pn TV = 2 2
1
1
1 1
3
torr)(300 m )(273 K)
solving for V2 ⇒ V2 = PPVTT ⇒ V2 = (597
(760 torr)(65 + 273 K)
V2 = 190 m3
1
1 2
n2T2
2 1
3. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne has a total pressure of 7.0 atm. What is the partial
pressure of Ne?
a. 1.4 atm
b. 2.2 atm
c. 3.8 atm
d. 4.6 atm
e. 2.7 atm
PNe = XNePtotal
XNe = nn
Ne
total
XNe = 3.5mol
11mol = 0.318
PNe = (0.318)(7.0 atm) = 2.2 atm
4. A mixture of oxygen and helium is 92.3% by mass oxygen. What is the partial pressure of oxygen if the total
pressure is 745 Torr?
If you had a 100 g sample ⇒ 92.3 g of O2 and 7.7 g of He
92.3 g O
7.7 g of He
= 2.88 mol O2
= 1.92 mol He
31.998 g/mol
4.0026 g/mol
2
2.88 mol O
PO2 = XO2Ptotal = (4.80
)745 torr = 447 torr
total mol
2
5. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr. How many oxygen molecules are in the
sample? a. 1.16 × 1023 b. 5.8 × 1022 c. 2.32 × 1024 d. 1.16 × 1022 e. none of these
# of molecules can be derived from moles
(800torr)(4.5L)
PV = nRT
n = 𝑃𝑉
n = (62.37torrL/molK)(300K)
n = 0.192 mol
𝑅𝑇
23
23
0.192 mol x 6.022x10 molecules/mol =1.16x10 molecules
6. Consider the combustion of liquid hexane: 2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l)
If 1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr. The carbon dioxide gas is collected
and isolated at 297 K and 0.930 atm. What volume of carbon dioxide gas will be collected, assuming 100%
yield? a. 0.504 L b. 1.93 L c. 2.23 L d. 0.607 L e. 4.04 L
Need to determine the limiting reagent
g
nC6H14 = 86.071.52g/𝑚𝑜𝑙
= 0.0177 mol vs.
(890 torr/760 torr/atm)(2.95L)
nO2 = PV
= 0.135 mol
RT (0.08206 atmL/molK)(312K)
0.135/19 < 0.0177/2 so O2 is the LR
mol CO )
0.135 mol O2 (12
= 0.0853 mol CO2
(19 molO )
2
2
atmL/molK)(297K)
V = nRT
= (0.0853 mol)(0.08206
= 2.23 L
P
(0.93atm)
7. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an evacuated cylinder with a volume of 1.60
L. The salt decomposes when heated, according to the following equation:
2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g)
Assuming complete decomposition, what is the pressure (in atm) in the cylinder after decomposition and cooling
to a temperature of 300. K? Assume the PbO(s) takes up negligible volume.
The reaction produces 2 gases so the pressure in the container is the total pressure ⇒ Ptotal = n VRT
total
(1mol Pb(NO3)2 )
(331 g Pb(NO3)2)
3.54 g Pb(NO3)2
NO +1mol O )
𝑥(4 mol
= 0.0276 mol gas
2 mol Pb(NO )
2
2
3 2
Ptotal = (0.0267 mol)(0.08206atmL/molK)(300K)
= 0.41 atm
(1.6L)
8. 2.5 mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a 3.50-liter container at 23°C. If the carbon and
oxygen react completely to form CO (g), what will be the final pressure (in atm) in the container at 23°C?
2 C (s) + O2 (g)  2 CO (g)
Determine limiting reagent
C ⇒ 3 mol C/2 = 1.5
O2 ⇒ 2.5 mol/1 = 2.5 ⇒ C is the LR
Since C is the LR ⇒ in addition to the CO formed there will be excess O2 in the container so the pressure
will be the total pressure ⇒ Ptotal = ntotalRT/V
Since there is 4 mol of gas in the container
0.08206𝑎𝑡𝑚𝐿
(4 𝑚𝑜𝑙)(
)(23+273𝐾)
𝑚𝑜𝑙𝐾
Ptotal =
3.5 𝐿
Ptotal = 27.8 atm
9. The density of an unknown gas at STP is 0.715 g/L. Identify the gas.
a. NO
b. Ne
c. CH4
d. O2
Molar mass can be useful to identify a substance
M = DRT
P
M = (0.715 g/L)(0.08206atmL/molK)(273K)
(1atm)
M = 16 g/mol
Unknown gas is CH4
10. Air is 79% N2 and 21% O2 by volume. Calculate the density of air at 1.0 atm, 25°C.
a. 0.590 g/L
b. 1.18 g/L
c. 2.46 g/L
d. 14.1 g/L
e. none of these
𝑚𝑎𝑠𝑠
Density =
𝑣𝑜𝑙𝑢𝑚𝑒
If there was 100 L of air there would be 79 L of N2 and 21 L of O2
Use PV = nRT to get moles
N2 ⇒ (1atm)(79 L) = n(0.08206 atmL/molK)(298K) ⇒ n = 3.23 mole N2
O2 ⇒ (1atm)(21 L) = n(0.08206 atmL/molK)(298K) ⇒ n = 0.859 mole O
11. These plots represent the speed distribution for 1.0 L of oxygen at 300 K and 1000 K. Identify which temperature
goes with each plot.
According to the average speed equation ⇒ 𝑈𝑎𝑣𝑒 = √8𝑅𝑇
𝜋𝑀
we can see the relationship between average speed and temperature
as T ↑ uavg ↑ since uavg B > uavg A ⇒ TB>TA
Plot A ⇒ 300K and Plot B ⇒ 1000K
12. The above plots represent the speed distribution for 1.0 L of He at 300 K and 1.0 L of Ar at 300 K. Identify which
gas goes with each plot. According to the average speed equation ⇒ 𝑈𝑎𝑣𝑒 = √8𝑅𝑇
we can see the relationship
𝜋𝑀
between average speed and molar mass ⇒ as molar mass ↑ uavg ↓ ⇒ since uavg B > uavg A ⇒ MB < MA
Plot A ⇒ Ar and Plot B ⇒ He
13. Calculate the temperature at which the average velocity of Ar (g) equals the average velocity of Ne (g) at 25°C.
⇒ √8𝑅𝑇𝐴𝑟
= √8𝑅𝑇𝑁𝑒
⇒ 8,R, and π constant
𝜋𝑀𝐴𝑟
𝜋𝑀𝑁𝑒
𝑢𝑎𝑣𝑒 = √8𝑅𝑇
⇒ uave Ar = uave Ne
𝜋𝑀
TAr
MAr
= MT ⇒ TAr =
Ne
Ne
MArTNe
MNe
g/mol)
⇒ TAr = (298K)(39.95
(20.18 g/mol)
⇒
T = 590 K or 317°C
14. It takes 12 seconds for 8 mL of hydrogen gas to effuse through a porous barrier at STP. How long will it take for the
same volume of carbon dioxide to effuse at STP?
timeCO2
timeH2
= √MM
CO2
H2
⇒ timeCO2 = timeH2 √MM
CO2
H2
⇒ timeCO2 = (12 s) √44.01g/mol
2.016g/mol
⇒ timeCO2 = 56 s
15. The diffusion rate of H2 gas is 6.45 times faster than that of a certain noble gas (both gases are at the same
temperature). What is the noble gas?
Molar mass can be used to identify ⇒
Munk = 2.016 g/mol √6.45
rateH2
rateunk
Munk
= √M
H2
rate
H2
⇒ Munk =MH2√(rate )
unk
⇒ Munk = 83.87 g/mol ⇒ Unknown gas is Kr
16. Consider two 5 L flasks filled with different gases. Flask A has carbon monoxide at 250 torr and 0 °C while flask B
has nitrogen at 500 torr and 0 °C.
a. Which flask has the molecules with the greatest average kinetic energy?
According to KEavg = 32RT ⇒ we see the relationship ⇒ as T ↑ KEavg↑ ⇒ since both flasks are at the same
T they will have the same KEavg
b. Which flask has the greatest collisions per second?
𝑍=
4𝑁𝑑2
𝑉
√𝜋𝑅𝑇
𝑀
we see three relationships ⇒ as T↑ Z↑ or as molar mass↑ Z↓ or as N/V (or P)↑ Z↑ ⇒ so since both flasks have the
same T and molar mass but the PB > PA ⇒ ZB > ZA
17. Under what conditions will a gas behave the most like an ideal gas?
An “ideal” gas is one that in reality adheres to the ideal gas law ⇒ meaning experimental values agree with
calculated values using PV = nRT
Gases are more likely to behave “idealy” when the pressure is low and/or the temperature is high
Deviations from the ideal gas law is due to the attractive forces between the gas particles and the volume of the
gas particles relative to the volume of the container
18. Which of the following gases will have the lowest molar volume at STP?
a. He
b. CH2Cl2
c. CO2
The molar volume of an “ideal” gas is 22.4 L/mol
as the attractive forces of the gas particles ↑ the molar volume ↓ – later (in ch 16)
we will learn the specifics of attractive forces however for now we can use the relationship that as molar
mass ↑ attractive forces ↑ (an exception is water – although water is rather on the light side it has quite
strong attractive forces called Hydrogen-Bonds which we’ll see further in ch 16)
Therefore since CH2Cl2 has the highest molar mass it has the strongest attractive forces and the lowest
molar volume
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