Grams of gasoline needed to raise level of CO in air.

[eddzzz_2011]

Homework Statement

Assume that an incorrectly adjusted lawn mower is operated in a garage such that the combustion reaction in the engion is C8H8. If the dimentions of the garage are 5x3x3 meters. How many grams of gasoline must be burned to raide the levl of CO in the air to 1000ppm by volume STP?

Homework Equations

PV=nRT

The Attempt at a Solution

101325 (45) = n(8.314)(298)
4559625 = n(2477.572)
n = 1840.36024moles

1840.36.../1000
1.84ppm

C8H8 + 6O2 ----> 8CO + 4H2O.

1 mole octane = 8 moles of CO

1.84/8 = 0.23 (moles of C8H8)

Mr C8H8 = 104.15 (have no idea why I took Mr of CO).

mass = n*Mr

mass = 0.23*104.15

mass = 23.98g gasoline needed

Is this correct? Cheers guys

 

[Borek]

1840.36.../1000
1.84ppm

1.84ppm? How is it related to the question?