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

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    PV=nRT

    3. 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
     
  2. jcsd
  3. Nov 2, 2011 #2

    Borek

    User Avatar

    Staff: Mentor

    1.84ppm? How is it related to the question?
     
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