Chemistry 110 Limiting Reactant

Question

C8H18(g)+O2(g)→CO2(g)+H2O(g)

how much octane is left if 0.110 mol of octane is allowed to react with 0.890 mol of oxygen.

Answer 1

Balanced equation

2C8H18 + 25O₂ ==> 16CO₂ + 18H2O

Moles octane used up = 0.890 mol O₂ x 2 mol octane/25 mol O2 = 0.0712 moles octane used in this reaction

moles octane left = 0.110 mol - 0.0712 mol = 0.0388 moles left

mass octane left = 0.0388 mol x 114g/mol = 4.42 g