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A limiting Reagent Problem: Check My work!

  1. Oct 29, 2006 #1
    Problem: combustion of propane, C3H8

    C3H8 (g) + 5O2 (g) --- 3 CO2 (g) + 4H2O (l)

    if propane reacts with oxygen as above
    (a) what is the limiting reagent in a mixture containing 5.00 g of C3H8 and 10.0 g of C3H8 of O2?
    (b) what mass of CO2 is formed when 1.00 g of C3H8 racts completely?

    My Approach:

    (a) 5.00g C3H8 x (1mol C3H8/ 44g C3H8) = .114 mol C3H8
    10.0g O2 x (1mol C3H8/ 32g O2) = .313 mol O2

    C3H8 is limiting

    (b) 1.00g C3H8 x (1 mol C3H8/ 44g C3H8) x (3mol CO2/ 1mol C3H8) x (44g CO2/ 1 mol CO2) = 3.00g CO2

    I'm kinda doubting my answer on the limiting reagent part...can someone explain it to me if im wrong?
  2. jcsd
  3. Oct 29, 2006 #2
    bump anyone?
  4. Oct 29, 2006 #3
    convert to moles first, and then divide by the coefficients. the smaller number indicates which one is the limiting reactant.

    So [tex] O_{2} [/tex] is the limiting reactant
    Last edited: Oct 29, 2006
  5. Oct 29, 2006 #4
    great tip thanks!
    they skip that kinda approach in college...it's all memorization!! i hate it...
  6. Oct 30, 2006 #5


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