Answer to Question #268111 in Chemistry for Jazz

Question

When 11.8 of C8H18 underwent a combustion reaction with 8.84g of O2 it produced 1.15g of H2O. Balance the reaction and answer the following questions. What is the limiting reactant?

Answer 1

2C₈H₁₈ + 25O₂ = 18H₂O + 16CO₂

n = m / M

M (H₂O) = 18 g/mol

M (C₈H₁₈) = 114.2 g/mol

M (O₂) = 32 g/mol

n (H₂O) = 1.15 / 18 = 0.064 mol

According to the reaction, in order to form this amount of water the required amount of incoming reactants is:

n (C₈H₁₈)_required = 2/18 x n (H₂O) = 2/18 x 0.064 = 0.007 mol

n (O₂)_required = 25/18 x n (H₂O) = 25/18 x 0.064 = 0.089 mol

The actual amount of incoming reactants is:

n (C₈H₁₈)_actual = 11.8/114.2 = 0.1 mol

n (O₂)_actual = 8.84 / 32 = 0.28 mol

Due to this, both incoming reactants are in excess.