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50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3 of O2. After explosion, the volume measured at r.t.p. was 37.0 cm^3. After treatment with aqueous KOH, the volume was reduced to 5.0cm^3. Calculate the % composition by volume of the original mixture.

    2. Relevant equations

    3. The attempt at a solution
    From this, I can safely assume that CO will react with O2 to form CO2
    2CO + O2 --> 2CO2
    2H2 + O2 --> 2H2O
    and CO2 will not react with Oxygen.
    When treated with KOH, the steam and carbon dioxide will all be removed. I assumed that 5.0cm^3 of oxygen is left. However, what I don't understand is how did the total volume of gas decrease from 75cm^3 (including oxygen) to 37.0cm^3?
  2. jcsd
  3. Feb 8, 2010 #2


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    Re: combustion

    Is there any steam at r. t. p.? Water vapor? How much water vapor will there be after treatment with KOH? More? Less? The same? I assume that the KOH treatment is an aqueous solution of KOH but your instructor might have a different idea.

    Three gases at the beginning. How many gases at the end (at r. t. p.)?
  4. Feb 9, 2010 #3
    Re: combustion

    Hmm. Good question. Is there steam? If there isn't, then everything becomes complicated.
    I guess the 'steam' will just stay with the KOH after treatment so that oxygen is the only gas that is left.
    So, at the end, there is oxygen.
    But I'm not that sure about that. Maybe not all carbon monoxide or hydrogen is reacted, so they may be collected? But, if I think it like that, I'm just complicating things.
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