How Do I Calculate the Volume of Air Needed for Complete Combustion of Gasoline?

[Gil-H]

Homework Statement

What volume of air (T=25C, P=1atm) is required
for complete combustion of one litter of gasoline?
The partial pressure of oxygen in the air is 0.205 atm.
One litter of gasoline contains 6.15 moles octane C8H18.

Homework Equations

The unbalanced reaction is:
C8H18 + O2 --> CO2 + H2O
The ideal gas law:
PV = nRT

The Attempt at a Solution

The balanced reaction is:
C8H18 + (25/2)O2 --> 8CO2 + 9H2O
So 6.15 moles octane requires 6.15*12.5 = 76.875 moles O2.
With the ideal gas law I get:
PV = nRT
(0.205)V=(76.875)(0.082)(298)
V = 9,163.5 L

Is this the correct answer?
My friend beleives it is,
but I think that this value is just the volume of oxygen needed,
and the volume of air needed is V = (9,163.5/0.2) = 45,817.5 L

 

[Borek]

Your friend is right.

Your approach would be correct if you would use not partial pressure of the oxygen, but 1 atm.

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methods

 

[Blueshift5]

I would say that your friend is correct. The calculation you have done gives the volume of oxygen needed for complete combustion of gasoline, but the question asks for the volume of air needed. This means that you need to take into account the fact that air is a mixture of gases, and only about 20.5% of the air is oxygen. Therefore, the volume of air needed would be about 5 times the volume of oxygen calculated, which is approximately 45,817.5 L. It is important to be clear about what the question is asking for and to consider all relevant factors when calculating a solution.