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Calculating total pressure

  1. Mar 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Sulfur dioxide gas reacts with oxygen to produce sulfur trioxide gas. Two 2 liter flasks at 25 degrees Celsius are connected by a stopcock. One flask contains 100 grams of sulfur dioxide, and the other with oxygen to a pressure of 5 atm. When the stopcock is opened the gases react until one is completely consumed. What will be the final pressure of the system after the reaction is completed.

    2. Relevant equations
    PV=nRT

    3. The attempt at a solution
    I started with the balanced equation: SO2 (g) + O2 (g) = SO3 (g). I then used stoichiometry to calculate the number of moles of SO2 in the first flask, and then calculated the number of moles of O2 in the second flask using PV=nRT. I'm pretty sure that this first part is right, but if not please advise me on what to do. After this is what I'm unsure of what to do. I wasn't sure if I was supposed to find the pressure of the sum of the number of moles of the two gases using PV=nRT or to subtract the smaller number of moles of O2 gas from the larger number of moles from the SO2 gas and then use PV=nRT (because the problem statement says that one gas will be completely consumed).
     
  2. jcsd
  3. Mar 7, 2015 #2

    SteamKing

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    Is this equation balanced?

    SO2 (g) + O2 (g) = SO3 (g)
     
  4. Mar 7, 2015 #3

    DrClaude

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    That's not balanced.

    What is left after the reaction is over? Then use that to calculate the total pressure.
     
  5. Mar 7, 2015 #4
    Oh I forgot to balance it.
    2SO2 (g) + 1O2 (g) = 2SO3 (g)
    That's the equation, thanks didn't catch that.
    So should I subtract the moles and then use PV=nRT or use the sum of the moles in PV=nRT. What's left is what I'm unsure of. Here I think since it's a limiting reactant and there's more moles of SO2 than O2 than the O2 becomes completely used up.
     
  6. Mar 7, 2015 #5

    DrClaude

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    I'll ask again: what do you have in the flasks after the reaction is finished?
     
  7. Mar 7, 2015 #6

    SteamKing

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    It's customary to omit the "1" in front of the O2. A quick glance and it seems you've written "102" instead. :rolleyes:

    You wouldn't write 1x + 2y = 3; the "1" is understood.
     
  8. Mar 7, 2015 #7
    The SO3 is the product and what's left. I understand now. Seems like I forgot how to do limiting reactant problems and had to look back on some old problems. I did my stoichiometry wrong also because I didn't account the moles of SO3 into this...
     
  9. Mar 7, 2015 #8

    DrClaude

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    You mean that there is only SO3 in the flasks at the end?
     
  10. Mar 7, 2015 #9
    Would it be the moles of SO3 formed and the moles of the excess reactant, SO2?
     
  11. Mar 7, 2015 #10

    SteamKing

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    How do you know which reactant, SO2 or O2, is in excess?
     
  12. Mar 7, 2015 #11
    So I worked out the problem. Using the number of moles of O2 I found from using PV=nRT, I found that 0.818 moles of SO3 would be formed. Using stoich. to find the number of moles of SO2 and then using those number of moles, I found that 0.781 moles of SO3 would be formed. That means that SO2 is actually the limiting reactant. So in calculating the total pressure I would need 0.781 moles of SO3 in addition to the moles of the excess reactant, correct? Now that means the O2 is the excess reactant. The number of moles of O2 before the reaction is 0.409 moles of O2 and the number of moles of SO2 before the reaction is 1.56 moles. The difference between these two numbers is 1.15 moles. Would this be the excess number of moles? If so would I then take 1.15 moles and add it with 0.781 moles for the total number of moles to plug into PV=nRT?
     
    Last edited: Mar 7, 2015
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