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Higher - Proportional Gas Volumes Q

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  • Message 1. 

    Posted by MissChemTeacher_Bitesize (U15172316) on Sunday, 18th November 2012

    Hey Guys!

    Here is a question on proportional gas volumes.

    A mixture of 20cm³ of propane and 130cm³ of oxygen was ignited and allowed to cool. Calculate the volume and composition of the resulting gas mixtures. (All volumes are measured at room temperature and at the same pressure)

    C₃H₆(g) + 5O₂(g) ----------> 3CO₂(g) + 4H₂O(l)


    Post your answers on the message board.

    I'll put up a solution in a few days.

    Cheers Miss_Chem smiley - cool

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  • Message 2

    , in reply to message 1.

    Posted by MissChemTeacher_Bitesize (U15172316) on Sunday, 25th November 2012

    Balanced Equation: C₃H₆(g) + 5O₂(g) ----------> 3CO₂(g) + 4H₂O(l)
    Mole ratios:……………1……..........5……………............3….......…..neg

    From the question we have 20cm³ of our fuel. We need to use this information to calculate what volume of oxygen will be needed for complete combustion as well as how much carbon dioxide will be released.


    Balanced Equation: C₃H₆(g) + 5O₂(g) ----------> 3CO₂(g) + 4H₂O(l)
    Data from the Q:…..20cm³…..100cm³……….....…....60cm³……..neg

    Our resulting gas volume will contain 30cm³ of un-reacted oxygen as well as 60cm³ of carbon dioxide.

    Hope this helps you out. Cheers Miss_Chem smiley - smiley

    Report message2

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