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

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To determine the grams of gasoline needed to raise CO levels to 1000 ppm in a garage measuring 5x3x3 meters, the combustion reaction of C8H8 (octane) is analyzed. Using the ideal gas law, the required moles of CO are calculated, resulting in approximately 1840.36 moles. This translates to 0.23 moles of C8H8, which, when multiplied by its molar mass of 104.15 g/mol, yields about 23.98 grams of gasoline needed. The calculation process raises questions about the relevance of the 1.84 ppm figure to the original query. The final answer suggests that approximately 24 grams of gasoline is required to achieve the desired CO concentration.
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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
 
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eddzzz_2011 said:
1840.36.../1000
1.84ppm

1.84ppm? How is it related to the question?
 

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