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Thermo Problem, desperate

  1. Apr 24, 2007 #1
    I have this reaction:

    CH4+4(O2+3.76) --> CO2+2H2O+2O2+15.04N2

    Through other calculations and given information I have found that the mass flow rate of the air (the O2+3.76N2) is 25kg/s and the mass flow rate of the fuel is .87 kg/s. I am having the hardest time finding, using these mass flow rates, the number of moles of each product I have. Any help is appreciated
  2. jcsd
  3. Apr 24, 2007 #2


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    Mass flow rates are given in kg/second, and the similarly you would get moles/second when converted into moles.

    How would you normally convert 1kg of O2 into moles of O2?

    Hint: 1kg/sec x moles/kg = moles/sec
  4. Apr 25, 2007 #3
    So if I have 25 kg/s of air and air is 21% O2 I have (.21*25kg/s) gives kg/s of O2. Then (.21*25 kg/s)/32 = N_O2? Similarly ((.79*25kg/s)(3.76N2))/28 gives N_N2

    How would it work for CO2 for example? Do I find the amount of moles of C, then use the fact that 25% of the moles of oxygen are paired with the C?
  5. Apr 25, 2007 #4


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    Get the kilograms/sec of each molecule using the mass percentages. Then convert it into moles/sec using that substance's molar mass.

    For example the molecule CO2 has a molar mass of (C)12.01+(O)16.00+(O)16.00=44.01 grams per mole.
  6. Apr 25, 2007 #5
    Ok this is what I got (I really appreciate the help by the way):
    CO2=44 g/mole
    H2O=18 g/mole
    O2=32 g/mole
    N2=28 g/mole

    Total mass of exhaust=565/12

    Mass flow of exhaust=mass flow air + mass flow fuel = 25.87 so i take the mass fractions from above and multiply them by that and they give:

    N2=19.27 kg/s

    And I know for example 1 mole of CO2=44 grams so 2.00 kg/s = 45.45 moles of CO2/s? Is this all correct?
    Last edited: Apr 25, 2007
  7. Apr 25, 2007 #6


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    Just make sure you carefully check each of your mass flow results for each substance (as I haven't checked them). The concept is correct, and yes 2kg/s of CO2 is equivalent to approx 45 moles/s of CO2. =)
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