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Homework Help: Complete Combustion - which hydrocarbon will consume more O2

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known data
    one gram of each of the following gases is introduced into a 10 L container at 25 degrees C

    a) propane
    b) ethane
    c) methane
    d) pentane

    which gas will consume the greatest mass of oxygen upon complete combustion?

    The solutions says that the right answer is D. I do not know why
    2. Relevant equations
    no equations. just chemical ones.

    3. The attempt at a solution

    My logic is that if I have greater number of moles of hydrocarbon, I will need more oxygen to consume.

    I thus thought that since methane has a lesser molar mass, we have greater moles, so we will need greater moles [which leads to a higher mass] of O2 gas needed.

    Could anyone please advise on how to solve this?

  2. jcsd
  3. Mar 31, 2016 #2


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    2017 Award

    Start with writing out balanced chemical equations for the complete combustion of each of these hydrocarbons.
  4. Mar 31, 2016 #3

    My equations are as following
    • propane: C3H8 + 5O2 --> 3CO2 + 4H2O
    • ethane: 2C2H6 + 7O2 ----> 4CO2 + 6H2O
    • CH4 + 2O2 ---> Co2 + 2H2O
    • C5H12 + 8O2 ----> 5CO2 + 6H2O
    I believe these equations are correct. right?

    I just did some stoichiometry with approximate values, and I got 0.111 moles of O2 needed to combust the hydrocarbon in answer D.

    My other calculated numbers are:
    a) 0.1136 mol O2 needed
    b) 0.1167 mol o2 needed
    c) 0.125 mol o2 needed
    d) 0.111 mol o2 needed.

    So far, it seems that the hydrocarbon in answer d does not have the maximum need for O2 moles.

    Could I have made an algebraic error?

    For example, a sample calculation for my number for part A is:

    (1 g C3H8)/(44 g C3H8) * ( 5 mol O2 / 1 mol C3H8) = 0.1136.

    I could be wrong.

    Please advise and thanks for your help!
  5. Mar 31, 2016 #4


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    I did the calculations as well and got the same numbers as you. The answer should be C (methane), and the book is probably wrong.
  6. Mar 31, 2016 #5
    Wow. This seems really strange.

    Thanks so much for the help!
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