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Determining the quantity in moles of the gas that remains unreacted ?

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the following reaction, which takes place in an autoclave at 250 degrees C and 800 atm.

    NH3(g) + 7/4 O2(g) --> NO2(g) + 3/2 H20(g)

    Into the reaction vessel has been placed 200 L of NH3(g) and 120 L of O2(g). The reaction is allowed to go to completion. Determine the quantity, in moles, of the gas that remains unreacted.


    2. Relevant equations



    3. The attempt at a solution

    I really don't know what to do, so I just tried this:
    The ratio of moles of O2 to NH3 is 7/4:1. Therefore the volume ratio is also 7/4:1. If you react 120 L of O2, then:
    120 x 4/7 = 68.6 L NH3

    200 L NH3 - 68.6 L NH3 = 131.4 L NH3 left unreacted

    Then I put it into n = PV/RT to convert to moles and I get 2448.2 mol NH3 left unreacted.

    IS THIS CORRECT??? Thanks
    Detailed explanations would be greatly appreciated.
     
  2. jcsd
  3. Mar 2, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Seems OK.
     
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