Limiting Reactants

Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 5.7 g of octane is mixed with 32.6 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction.

Answer 1

Octane = C8H18 (MW = 114 g / mol)

Oxygen = O2 (MW = 32 g/mol)

Carbon dioxide = CO2 (MW = 44 g/mol)

Balanced equation:

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

Limiting reactant:

mols C8H18 = 5.7 g C8H18 x 1 mol / 114 g = 0.050 mols(÷2->0.025) This is LIMITING REACTANT

mols O2 = 32.6 g O2 x 1 mol / 32 g = 1.02 mols (÷25->0.04)

Since octane (C8H18) is determined to be the limiting reactant, the minimum mass that could be left over would be zero. It should all be used up, and O2 would be left over.