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Volume - Volume Calculations

  1. Dec 18, 2006 #1
    Suppose that you want to prepare 2.00*10^3 L of ammonia gas, NH3 , from the reaction of hydrogen gas with nitrogen gas. What volume of hydrogen and nitrogen do you need? Assume that the temperature and pressure remain constant during the reaction.

    3H2(g) + N2(g) -> 2 NH3(g)

    V H2 = 2.00*10^3 L NH3 * (3 vol H2 / 2 vol NH3) = 3.00*10^3 L H2

    V N2 = 2.00*10^3 L NH3 * (1 vol N2 / 2 vol NH3) = 1.00*10^3 L N2

    Therefore, 3.00*10^3 L of hydrogen and 1.00*10^3 L of nitrogen is required to make 2.00*10^3 L of ammonia.

    This is an example in my text book. My question is: how is it that when 3000L of hydrogen and 1000L of nitrogen are combined it produces only 2000L of ammonia?

    Is the question stated incorrectly, is it taking into account only one volume of ammonia produced, or does the chemical reaction between hydrogen and nitrogen produce a more closely condensed ammonia gas?
  2. jcsd
  3. Dec 18, 2006 #2


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    Your textbook is not wrong. If you look at the equation, four molecules on the left produce 2 molecules of ammonia. So, it makes sense that 4000L are required to produce 2000L of ammonia.
  4. Dec 18, 2006 #3
    So, does a mole of ammonia, take up the same physical space as either a mole of hydrogen or nitrogen would on their own? By combining the two is the total volume of gases reduced?

    3 vol H2 + 1 Vol N2 = 2 Vol NH3

    3L H2 + 1L N2 = 2L NH3
    Last edited: Dec 18, 2006
  5. Dec 18, 2006 #4


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    At standard temperature and pressure, one mole of any gas will occupy the same volume (~22L I think). So yes, if the reaction was to work completely, then you would be left with a yield of half the volume of the gases you put in.
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