Download 1 Stoichiometry Problems Volume of CO2 (g) produced from the

Stoichiometry Problems
Volume of CO2 (g) produced from the combustion of one gallon of gasoline:
CH 222
Octane to CO2 Notes
Oregon State University
Dr. Richard Nafshun
A couple of CH 222 students hop into a Toyota and drive to Eugene to visit an unenlightened friend. Using a
few opportune reaction conditions, determine the liters of CO2 produced. What is the length (in feet) of a cubic
container that would hold the CO2?
Molar Mass of Octane = 114.2285 g/mol
1.00000 Gallons = 3.78541 Liters
1.0000000 Liter = 0.03531467 cubic foot
Density of octane = 0.703 g/ml
Mileage Est. (mpg city/highway): 32/41
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a.
I will assume the combustion of 1.00 gallon of octane and the existence of
CO2 (g) at 298 and 1 atm when at equilibrium:
 3.785 L 
 = 3.785 L C8H18
1.00 gal C8H18 
 1.000 gal 
b.
 1000 mL 
 = 3785 mL C8H18
3.785 L C8H18 
 1L 
c.
 0.703 g 
 = 2660 g C8H18
3785 mL C8H18 
 1 mL 
1.
 1 mol 
 = 23.3 mol C8H18
2660 g C8H18 
 114.21 g 
2.
 8 mol CO 2 
 = 186 mol CO2
23.3 mol C8H18 
 1 mol C 8 H18 
d.
PV = nRT
L  atm


)(298 K ) 
 (186mol)(0.0821
nRT

 
mol  K
 = 4550 L CO2 (g)
V= 
 =
(1.00 atm)

 P  




2
[Change L to ft3 and discuss a cube that holds the carbon dioxide]:
 0.0353 cubic feet 
 = 161 ft3 CO2 (g)
4550 L CO2 (g) 
1L


Volume of a cube = side3
Side = 3√V = 3√161 ft3
Side = 5.44 ft (5’ 5”)
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"Special Cases" or "A gas undergoes a change" problems
For a balloon:
A gas at a temperature of 98.0 C (371 K) occupies a balloon with a volume of 125.0 mL. To what
temperature must the gas be lowered to occupy a volume of 100.0 mL?
PV = nRT
PV/nT = R = constant
P1V1
PV
= 2 2
n1T1
n 2T2
P is a constant
V is NOT a constant
n is a constant
V1
V
= 2
T1
T2
125.0 mL = 100.0 mL
371 K
T2
T2 = 296.8 K or 23.8 ºC
We must use K in
P1V1
PV
= 2 2
n1T1
n 2T2
Why?
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For a flask:
P1V1
PV
= 2 2
n1T1
n 2T2
P is a NOT a constant
V is a constant
n is a constant
P1
P
= 2
T1
T2
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Density of gases:
What is the density (g/L) of SF6 under standard conditions, 1.00 atm and 273K?
Assume one liter of gas (so the mass can be determined for one liter).
PV = nRT
n = PV/RT = (1.00 atm)(1.00 L)/(0.0821 L•atm/mol•K)(273 K) = 0.0446 moles SF6
 146.06 g 
0.0446 moles SF6 
 = 6.51 grams SF6
 1mole 
6.51 grams is the mass of one liter, so:
The density of SF6 is 6.51 g/L.
___
What is the density (g/L) of He under standard conditions, 1.00 atm and 273K?
Assume one liter of gas (so the mass can be determined for one liter).
PV = nRT
n = PV/RT = (1.00 atm)(1.00 L)/(0.0821 L•atm/mol•K)(273 K) = 0.0446 moles He
 4.002 g 
0.0446 moles SF6 
 = 0.178 grams He
 1mole 
0.178 grams is the mass of one liter, so:
The density of He is 0.178 g/L.
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Partial Pressure:
A 10.0-L flask at a temperature of 300.0 K contains one mol of O2 gas, 2 moles of CO2 gas, and 7 moles
of N2 gas. Calculate the total and partial pressures.
PV = nRT
P = nRT/V
The partial pressure of oxygen (the pressure inside the flask due to the oxygen gas molecules:
P(oxygen) = n(oxygen) RT/V
P(oxygen) = (1.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 2.46 atm
The partial pressure of carbon dioxide (the pressure inside the flask due to the carbon dioxide gas
molecules:
P(oxygen) = n(carbon dioxide) RT/V
P(oxygen) = (2.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 4.93 atm
The partial pressure of nitrogen (the pressure inside the flask due to the nitrogen gas molecules:
P(oxygen) = n(nitrogen) RT/V
P(oxygen) = (7.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 17.2 atm
Total Pressure = 2.46 atm + 4.93 atm + 17.2 atm = 24.6 atm
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Real Gases:
Recall for ideal gases:
PV=nRT
The van der Waals equation for real gases:
"a" accounts for the attractive forces between gas particles (a is low for small, non-polar, gases)
"b" accounts for the actual gas particle size (b is low for small gases)
van der Waals Coefficients
Gas
a (Pa m3) b(L/mol)
Neon
0.0212
0.1710
Hydrogen
0.0245
2.661
Carbon dioxide
0.396
4.269
Water vapor
0.547
3.052
Another, more sophisticated model:
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