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Ammonia reacts with O2 problem

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Ammonia reacts with O2 to form either NO(g) or NO2(g) according to these unbalanced equations:
    NH3 + O2 ----> NO + H2O
    NH3 + O2 ----> NO2 + H2O
    In a certain experiment 2.00 moles of NH3 and 10.00 moles
    of O2 are contained in a closed flask. After the reaction is
    complete, 6.75 moles of O2 remains. Calculate the number
    of moles of NO in the product mixture: (Hint: You cannot
    do this problem by adding the balanced equations because you
    cannot assume that the two reactions will occur with equal
    probability.)

    2. Relevant equations

    3. The attempt at a solution
    After the balance of equations,
    x=mol of NH3 in the first equations(=mol of NO as well)

    1st reaction:
    2 mol NH3 gives 5/2 mole O2
    x mol gives ... 5x/4 mole O2

    2nd reaction:
    2 mol NH3 gives 7/2 mole O2
    (2-x) mol gives ... 7(2-x)/4 mole O2

    5x/4 + 7(2-x)/4 + 6.75=10
    I find: x=0.5 mole of NH3 as well as 0.5 mole of NO
    But solution manual has 0.48 as answer, is my calculation wrong in any way??
     
  2. jcsd
  3. Apr 20, 2015 #2

    Borek

    User Avatar

    Staff: Mentor

    Any particular reason to not write balanced reaction equations and to make us guess whether you had them right or wrong?

    Ammonia doesn't "give" oxygen, it reacts with oxygen consuming it.

    But in general 0.5 moles of NO look correct.
     
  4. Apr 20, 2015 #3
    No, no particular reason :)

    2 NH3 + 5/2 O2 ----> 2NO + 3H2O
    2 NH3 + 7/2 O2 ----> 2NO2 + 3H2O
     
  5. Apr 20, 2015 #4
    Thank you both! :)
     
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