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Material balance with a reaction (chemical engineering)

  1. Jul 28, 2016 #1
    1. The problem statement, all variables and given/known data

    Felder & Rousseau 4.73 (p. 184)

    A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole % H2O. After all of the water is removed from the products, the residual gas contains 69.4 mole% CO2 and the balance is O2. What is the mole percent of propane in the fuel?

    2. Relevant equations

    1. Let propane = P
    2. Choose 100 mol of the reactant fuel as a basis

    3. The attempt at a solution

    Set up 2 balanced equations: 1. C3H8 + 5O2 -> 4H2O + 3CO2
    2. C4H8 + 13/2O2 -> 5H2O + 4CO2

    I assumed a 100% conversion of the reactant gases. I know O2 is in excess. I drew up a reactant table in terms of mol and set Propane in = P and Butane in= 100-P. I then tried to calculate P by setting the mols of water out divided by the total mols out equal to 0.474.
     
  2. jcsd
  3. Jul 28, 2016 #2
    So, what's the problem?
     
  4. Jul 28, 2016 #3

    collinsmark

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    Hello @MickeyBlue! Welcome to Physics Forums! :smile:

    :welcome:

    So far so good, with one exception:

    You wrote that the butane was [itex] \mathrm{C_4 H_8} [/itex], but I think you meant [itex] \mathrm{C_4 H_{10}} [/itex]. Your equations were balanced correctly though, so I assume that was just a minor mistake typing things into the computer.

    As a hint for moving forward, don't forget to include the leftover [itex] \mathrm{O_2} [/itex] in your product. Similar to what you did with [itex] P [/itex], make up a variable name to indicate the amount of [itex] \mathrm{O_2} [/itex] in there. I used the variable [itex] x [/itex], but you can use whatever you want.

    So your formula, as you have already stated it, should be of this form:

    [tex] \frac {\mathrm{moles \ of \ H_2O}}{(\mathrm{moles \ of \ H_2O}) + (\mathrm{moles \ of \ CO_2}) + (\mathrm{moles \ of \ O_2})} = 0.474[/tex]

    After that, take out the [itex] \mathrm{H_2O} [/itex] and do something quite similar except with [itex] \mathrm{CO_2} [/itex] in the numerator (and the ratio value being different).

    If you do things right, you'll have two equations and two unknowns ([itex] P [/itex] and [itex] x [/itex]). :wink: That's enough to solve for [itex] P [/itex].
     
  5. Jul 30, 2016 #4
    My final answer for propane was well above 100 and I couldn't understand why.
     
  6. Jul 30, 2016 #5

    collinsmark

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    Show your work, and perhaps we may be able to point you in the right direction.
     
  7. Jul 30, 2016 #6
    Oh! Thank you, I see where I went wrong now. Because of how I added mols of O2 consumed I took it as the amount in excess instead. You're right; this should be enough to get P.
     
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