2C8H18 + 25O2 -> 16CO2 + 18H2O

"0.170 mol of octane is allowed to react with 0.750 mol of oxygen. Which is the limiting reactant?"

the answer is oxygen, but can someone show me mathematically how it's oxygen please? thank you

9 months ago 2


  1. pisgahchemist

    Limiting reactant problem.....

    One way to deal with limiting reactant problems in order to find the limiting reactant is to assume that one reactant is the limiting reactant and find the amount of the other reactant which reacts. Then compare this value to the actual amount of the second reactant to see if your original assumption is correct. From that you can determine the limiting reactant.

    2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)
    0.170 mol .... 0.750 mol

    Assume octane is the limiting reactant. Compute the moles of O2 which will react:
    0.170 mol C8H18 x (25 mol O2 / 2 mol C8H18) = 2.125 mol O2

    You can see that for all of the octane to react requires 2.125 mol O2, but only 0.750 mol O2 is available, making O2 the limiting reactant.

  2. Roger the Mole

    0.750 mole of O2 would react completely with 0.750 x (2/25) = 0.0600 mole of C8H18, but there is more C8H18 present than that, so C8H18 is in excess and O2 is the limiting reactant.

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