Answer to Question #115915 in General Chemistry for lauren

Question

Suppose 9.1 g of octane is mixed with 59.5 g of oxygen. Calculate the minimum mass of octane that could be leftover by the chemical reaction

Answer 1

C₈H₁₈ + 12.5O₂ = 8CO₂ + 9H₂O

n(C8H18) = ${\frac {9.1}{114}}=0.08$ mol

n(O2) = ${\frac {59.5}{32}}=1.86$ mol

the oxygen is taken in excess, so octane itself is a limiting reactant, for 0.08 moles of octane 1 mol of oxygen is needed for complete combustion. So all octane will be burnt, and 0.86 mol of oxygen is left.