- #1
eddzzz_2011
- 17
- 0
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