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Homework Help: Parameterization Question

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Ethylene oxide is produced industrially from the reaction of ethylene with oxygen at atmospheric pressure and 283 oC, in the presence of silver catalyst.

    [tex]C_{2}H_{4} + O_{2} \rightarrow C_{2}H_{4}O[/tex]

    Assuming 100 % yield, how many kg of ethylene oxide can be produced from 35100 L of a mixture containing ethylene and oxygen in 1:1 molar ratio?

    2. Relevant equations

    3. The attempt at a solution

    First I balanced the equation like so,

    [tex]2C_{2}H_{4} + O_{2} \rightarrow 2C_{2}H_{4}O[/tex]

    I then converted the temp. from celsius to kelvin so,


    I'm not sure what value I should use for pressure, in a question like this am I just assuming that P = 1atm?

    I then applied PV= nRT and solved for moles. Once I have the number of moles I used the molar mass to get it into grams and then I converted grams to kilograms.

    I still got the answer wrong however.

    What am I doing wrong?

    EDIT:Also, what value for the gas constant do I use that will work with the units K, mol, atm, and L?
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 30, 2010 #2


    User Avatar

    Staff: Mentor

    Show details of your work.

    You are told it is happening at atmospheric pressure, and that means 1 atm.

    R value - check whichever fits the units. There are many lists of R values on the web.
  4. Sep 30, 2010 #3
    [tex]n = \frac{PV}{RT}[/tex]

    [tex]n = \frac{1atm \cdot 35100L}{0.082 \cdot 525.15K} = 815.10 \text{mols of Ethylene oxide} [/tex]

    Molar mass of Ethylene oxide 44.05 g/mol


    [tex]44.05\text{g/mol} \cdot 815.10\text{mols} = 35905.07g = 35.91kg[/tex]

    This answer is incorrect.

    What am I doing wrong?
    Last edited: Sep 30, 2010
  5. Sep 30, 2010 #4


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    Staff: Mentor

    Think about initial amounts of gases and stoichiometry.

    And check your kelvins.

  6. Sep 30, 2010 #5
    Oh, the temperature should be 556.15K, my bad!

    Can I get another nudge? It's still not obvious to me what to do next.
  7. Oct 1, 2010 #6


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    Staff: Mentor

    Do you have enough oxygen?
  8. Oct 1, 2010 #7
    So I should use PV=nRT to find the mols of oxygen, then I'll know whether it's a limiting reactant or not, correct?
  9. Oct 1, 2010 #8


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    Staff: Mentor

    Yes and no. Yes - you have to check what is a limiting reagent. No - you don't need PV=nRT for that. You are already told what is mixture composition.

  10. Oct 2, 2010 #9
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