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Homework Help: Mole Calculation

  1. Jul 20, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    An incomplete reaction of 12.125 g of propane proceeds as follows:

    3 C3H8(g) + 14 O2(g) → 8 CO2(g) + 12 H2O(g) + C(s)

    How many grams of solid carbon residue are produced as a product?

    So, C3H8

    C=3(12.01)=36.03 g/mol.
    H=8(1.01) = 8.08 g/mol.
    = 44.11 g/mol.

    Therefore, 12.125 g C3H8 x (1 mol. C3H8/44.11g C3H8) = 0.2749 mol C3H8

    Then, 0.2749 mol. C3H8 x (1 mol. C/3 mol. C3H8) = 0.09163 mol C

    So, 0.09163 mol. C x (12.01 g C/ 1 mol. C) = 1.101 g C?
  2. jcsd
  3. Jul 21, 2016 #2


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    Gold Member

    I get the same answer as you but couldn't follow your argument and suggest you won't be able to if you come back to it after 6 months. If people can't trace a simple logical process, you won't get the credit if you make a mistake, nor will you so easily detect a mistake.

    I would just set it out 3 moles propane gives 1 mole C;

    Therefore (multiplying by molecular masses) so many g propane gives so many g C

    Therefore (simple proportions) the stated number of propane gives so many g C
  4. Jul 21, 2016 #3
    Thanks epenguin!! I'll definitely revamp my form.
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