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    please help, i just dont know how to approach this problem

    The question gives us this info
    Most petrol has a relative octane number of 95. This means that it behaves in the same way as a mixture of 95% 2,2,4-trimethylpentane and 5% heptane.


    then it asks us to write the equation for octane burning completely in oxgen which is okay... but thn it asks us to determine the fuel to air ratio by volume of the complete combustion of gaseous octane

    i know you have to convert the rmm to volume somehow but i just draw a blank, please help if you can

    Thanks xxxx
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    (Original post by b33jal)
    please help, i just dont know how to approach this problem

    The question gives us this info
    Most petrol has a relative octane number of 95. This means that it behaves in the same way as a mixture of 95% 2,2,4-trimethylpentane and 5% heptane.


    then it asks us to write the equation for octane burning completely in oxgen which is okay... but thn it asks us to determine the fuel to air ratio by volume of the complete combustion of gaseous octane

    i know you have to convert the rmm to volume somehow but i just draw a blank, please help if you can

    Thanks xxxx
    Hmm, I think you would say that 12.5 mol of oxygen react with 1 mol of octane. (That's right isn't it? that's just done in my head so confirm with your equation)
    Assume air is made of 20% oxygen, so 62.5 mol of air react with 1 mol of octane (assuming air is a mixture of ideal gasses)

    Because 1 mole of octane and 1 mole of air will occupy the same volume (as they are both ideal gasses), you can say that the ratio of volumes is 1:62.5 (fuel:air)
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    12.5 moles of oxygen do react with octane. that's right.

    How do you know 62.5 mol of oxygen/air reacts with one mol of octane?
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    (Original post by b33jal)
    12.5 moles of oxygen do react with octane. that's right.

    How do you know 62.5 mol of oxygen/air reacts with one mol of octane?
    because oxygen is 20% of air - by volume and therefore by moles as we assume that air is made up of ideal gasses, all of which take up the same volume for 1 mole.

    Therefore 12.5 mol of oxygen can be found in 12.5 x 5 moles of air (because oxygen is 1/5 of air).

    12.5 x 5 = 62.5
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    ohhhhh i get it!!

    CHEERS MATE!
 
 
 
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