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Find the molar percentage...

  1. Jan 15, 2017 #1
    1. The problem statement, all variables and given/known data

    The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water produced is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.34 atm when measured in the same volume and at the same temperature as the original mixture. What was the composition of the original mixture in mole percent?
    2. Relevant equations
    2H2 +O2 ---> 2H2O

    3. The attempt at a solution


    1atm - 0.34atm = 0.66atm H2O

    for every 1 mol of H2 that reacts, 2 mols of H2 react so,
    0.44atm of the water is H2
    0.22 atm of the water is O2

    0.34atm +0.44atm = 0.78atm H2O
    0.22atm O2

    78% of the mixture is H2
    22% of the mixture is O2

    Can someone check this?
     
  2. jcsd
  3. Jan 15, 2017 #2

    epenguin

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    Gold Member

    I think that is quite OK.

    However for this sort of calculaction if you haven't got anyone to check it for you you can do it yourself. You just take your conclusions and pose the inverse question, suppose you have this initial composition (which is your answer) how much H2 would you get from it if you reacted it? - the answer should correspond to the first question.
     
  4. Jan 15, 2017 #3

    BvU

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    Hello Jeremy, :welcome: (somewhat belated).

    PF isn't really in the business of stamp-approving answers -- it would only create tension between PF and teachers in general -- especially in cases where opinions differ. Read the guidelines ...

    However, in your case there is little room for doubt, uncertainty and fear. So I wonder why you ask at all ?
     
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