Answer to Question #302523 in Chemistry for Kimbop

Question

Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous Carbon dioxide (CO2) and gaseous water (H2O). Suppose 10. g of octane is mixed with 69.1g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction

Answer 1

2C₈H₁₈ + 25O₂ = 16CO₂ + 18H₂O

According to the equation, n (O₂) = 25 / 2 x n (C₈H₁₈).

n = m / M

M (C₈H₁₈) = 114.2 g/mol

M (O₂) = 32 g/mol

The available amounts of reactants

n (C₈H₁₈) = 10 / 114.2 = 0.088 mol

n (O₂) = 69.1 / 32 = 2.16 mol

C₈H₁₈ is the limiting reactant. The amount of O₂ left:

n (O₂)_left = 2.16 - (25/2 x 0.088) = 1.06 mol

m (O₂)_left = n x M = 1.06 x 32 = 34 g