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    Exam in january and I'm stessing out, finding gas calculations really difficult.

    The Q is:
    18g of pentane is completely burnt in a car engine to form c02 and water.

    What volume of 02 is needed, assuming that one mole occupies 24dm^3?
    What volume of air is needed (assume that air contains 20% 02 by volume)
    What volume of carbon dioxide is produced?

    Thank you so much!

    P.S. Please write down step-by-step method and perhaps explaining why a certain step is used

    Thanks again, much appreciated
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    First you need to work out the moles of pentane used. Which is 18 / ((12 X 5) + (1 X 12) = 0.25 moles of pentane used.

    Now work out a balance equation:

    C5H12 + 5O2 >>>>>> 5 CO2 + 5 H2O

    So that means for everyone one mole of pentane you need 5 moles of O2 thus:

    0.25 * 5 = 1.25 moles of O2 needed.

    Next you need to times 24dm^3 by the number of moles of O2 used (1.25) thus: 1.25 * 24 = 30dm^3 of O2 used.

    As you know 20% is 30dm^3 you need to do 30dm^3 / 20 * 100 = 150dm^3 of air needed.

    Finally the volume of CO2 produced is equal to the volume of O2 produce so 30dm^3

    I hope that makes sense, if not please do quote me and ask.
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    (Original post by joshed)
    Lots of answers
    It's better to help someone work out the answers, than give them the answers.
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    (Original post by gingerbreadman85)
    It's better to help someone work out the answers, than give them the answers.
    Which is why I gave them a step by step method to doing it, it is their choice if they take my answer or use my workings to understand it.
 
 
 
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