Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 years ago

A general formula for gasoline is C7H13. Calculate the air to gasoline mass ratio required for stoichiometric combustion.?

1. A general formula for gasoline is C7H13. Calculate the air to gasoline mass ratio required for stoichiometric combustion.

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  • 3 years ago
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    Chances are you mean heptane, C7H16. Heptane by itself is unacceptable as a fuel and is found in low percentages in gasoline because of the explosive way it burns.

    Heptane and iso-octane (2,2,4-trimethylpentane) are used to determine the octane rating of gasoline. Heptane creates engine knock when the fuel burns explosively. By mixing heptane and iso-octane engine knock is reduced. A fuel mixture under test is compared to a mixture of heptane and iso-octane which gives the same amount of engine knock. The percentage of iso-octane in the heptane/iso-octane mixture is the octane rating of the fuel being tested.

    The ideal ratio would occur when there is complete combustion of the fuel, which does not occur in an actual engine. Assume a 100.0 gram sample of "gasoline."

    C7H16(g) + 11O2(g) --> 7CO2(g) + 8H2O(g)

    100.0g ......... ?g

    100.0g C7H16 x (1 mol C7H16 / 100.0g C7H16) x (11 mol O2 / 1 mol C7H16) x (32.0g O2 / 1 mol O2) = 352.0g O2

    Air is 21% oxygen. 100g of air contains 21 g O2 (actually closer to 20.95%)

    352.0g O2 x (100.0g Air / 20.95g O2) = 1680.g Air

    Air to gasoline ratio = 1680.g air / 100g C7H16 = 16.80:1

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  • DrBob1
    Lv 7
    3 years ago

    An odd number of Hydrogens? No way.

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