What is the greatest amount of water in grams that can be produced 215 grams of C7H16 and 87 grams O2? Which reactant is limiting? How many grams of the excess reactant will be left?

Answer 1

First write a balanced combustion reaction for the mole ratios: C₇H₁₆ + 11 O₂ --> 7 CO₂ + 8 H₂O.

(215g C₇H₁₆) * (1 mole C₇H₁₆ /100.2g C₇H₁₆) * (8 mole H₂O /1 mole C₇H₁₆) = 17.1₇ mole H₂O.

(87g O₂) * (1 mole O₂ /32.00g O₂) * (8 mole H₂O /11 mole O₂) = 1.9₈ mole H₂O.

Oxygen is limiting, maximum mass of H₂O is: (1.9₈ mole H₂O) * (18.02g H₂O /1 mole H₂O) = 36g H₂O.

Excess reactant from mole difference: (Δmoles product = 15.2 mol H₂O) * (1 mole C₇H₁₆ /8 mole H₂O) * (100.2g C₇H₁₆ /1 mole C₇H₁₆) = 190. g C₇H₁₆.