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Partial pressures and PV = nRT

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Ammonia burns in air to form nitrogen dioxide and water

    4NH3(g) + 7O2(g) --> 4NO2(g) + 6H2O(l)

    If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? (assume constant temperature)


    3. The attempt at a solution

    This is what i did:
    i used PV = nRT.... since T, R, and V are constant here,
    I said Pinitial/ ninitial = Pfinal/nfinal

    I used 11 atm for P initial and I added the moles of the 2 reactant gases for the initial n to get 8+14 = 22.
    (is this allowed? Can I add the 2 moles of gases together for this problem even though they are different gases?)

    Then I used 8 moles for my final n (since 8 moles of NO2 were formed) and then calculated 4 atm as my answer for the final partial pressure of NO2.


    Is there a faster/ easier way to calculate the answer?
    Did I do it the right way?
     
  2. jcsd
  3. Jul 22, 2011 #2
    You can do that; you actually probably want to consider PtotalV = ntotalRT (total being the two gases you start with) and PNO2V = nNO2RT, then combine the two equations to get Ptotal/ntotal = PNO2/nNO2 (similar to what you did).

    Your initial amounts of NH3 and O2 are in the same ratio as their reaction ratios in the equation, and 8 mol NH3 will yield 8 mol NO2, so everything you did looks good. :smile:
     
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