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Using gas law to find moles of SO3 formed

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data
    A chemist studying the rxns of pollutant SO2 places a mixture of SO2 and O2 in a 2.0L container at 900K and an P[init]=1.95 atm. When rxn occurs, gaseous SO3 forms and P falls to 1.65 atm. How many moles of SO3 formed?


    2. Relevant equations
    PV=nRT but I'm not sure what else...


    3. The attempt at a solution
    I was assuming both SO2 and O2 in the starting material are gases? So I used V=2L, P=1.95atm, and T=900K to find the number of moles of both gases combined, which turned out to be ~0.053 moles. And I'm not sure what to do with this number to get to the answer... I wrote out the balanced equation to be: 2SO2(g) + O2(g) --> 2SO3 (g) although I don't know if that matters... I can't assume that T stays the same, can I? the problem doesn't say anything about it...

    [edit] wait, can I start out by saying that... since you're not losing any gas, #moles of starting material (SO2+O2) has to be the same as #moles of final mixture (SO3+(leftover)SO2+O2)... and so use 0.053moles, given final pressure (1.65atm) and 2L (since it's a container it doesn't expand..?) to find the final temperature and go from there somehow?
     
    Last edited: Jan 21, 2007
  2. jcsd
  3. Jan 21, 2007 #2

    Clausius2

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    Try something called Chemical Equilibrium.
     
  4. Jan 21, 2007 #3
    hmm....? :-/
     
  5. Jan 21, 2007 #4

    Gokul43201

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    No, you have to assume that the volume of the container and the temperature are unchanged. And the number of moles is different from the number of atoms. It is the total number of atoms that is conserved, not the total number of molecules (or moles).

    1. Find the final number of moles using the final pressure.
    2. Translate the balanced equation into words.
     
  6. Jan 21, 2007 #5

    Clausius2

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    Allright, don't try something called Chemical Equilibrium.
    Try something called Follow What Gokul Pointed Out.
    :biggrin:
     
  7. Jan 21, 2007 #6
    so I did P(i)/n(i)=P(f)/n(f)... and got the final #moles to be 0.0447moles of gas... but I'm not sure what you mean by "translate the balanced equation into words."... so from 3 moles of gas you make 2 moles of SO3 gas...?

    Im guessing you can't assume that not all of 0.0447 moles of what's produced is SO3 because you can't assume that you had the exact amount of starting material needed to produce that, so the product gas mixture is a combination of all 3 gases?
     
  8. Jan 21, 2007 #7

    Gokul43201

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    Correct. All that's produced is NOT SO3.

    This is close. You need to expand on this idea. So, for every 3 moles consumed, there are 2 moles produced - resulting in a net loss of 1 mole.

    Thus, 1 mole net loss => 2 moles SO2 produced.
    So (0.053-0.0447) moles net loss => ?? moles of SO2 produced.
     
    Last edited: Jan 22, 2007
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