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    Complete combustion of 40 cm3 of a gaseous hydrocarbon X requires 240 cm3 of oxygen. 160 cm3 of
    carbon dioxide forms. All gas volumes are at room temperature and pressure.What is the formula of X?

    Please tell me how to do this. Thank you
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    (Original post by Nina31)
    Complete combustion of 40 cm3 of a gaseous hydrocarbon X requires 240 cm3 of oxygen. 160 cm3 of
    carbon dioxide forms. All gas volumes are at room temperature and pressure.What is the formula of X?

    Please tell me how to do this. Thank you
    C4H8 is the formulae (I think) That is what I got when I worked it out.
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    (Original post by Nina31)
    Complete combustion of 40 cm3 of a gaseous hydrocarbon X requires 240 cm3 of oxygen. 160 cm3 of
    carbon dioxide forms. All gas volumes are at room temperature and pressure.What is the formula of X?

    Please tell me how to do this. Thank you
    Yes, so this is how I worked it out(if you think this is wrong pls correct me).

    lets write the equation; X+O2------>CO2+H20

    40CM3 of X reacts with 240cm3 of O2 to give 160cm3 of water. therefore the volume ratios are 1;6;4

    to produce 4 moles of CO2 we obviously know that the X contains 4 carbons. Now we need to know what H is because this can know either be an alkene or an alkane.

    so I did two combustion equations for c4h10 and c4h8 to see what would give me 6 moles of O2

    -----> c4h10+13/2o2----->4co2+5hro
    ------>c4h8+6o2------->4co2+4h2o

    therefore X is c4h8
 
 
 
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