Liquid octane (CH3(CH2)6 (CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O).

Question

If 8.68g of water is produced from the reaction of 29.7g of octane and 43.7g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

= ____ %

Answer 1

Octane (C₈H₁₈)

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O ... balanced equation

29.7g........43.7g........................8.68 g

Find the limiting reactant (divide mols of each by corresponding coefficient)

29.7 g C₈H₁₈ x 1 mol C₈H₁₈ / 114 g = 0.261 mols C₈H₁₈ (÷2->0.13)

43.7 g O₂ x 1 mol O₂ / 32 g = 1.37 mols O₂ (÷25->0.05)

O2 is limiting since 0.05 is less than 0.13

Moles of O2 will determine mols of H₂O that can form.

Theoretical yield of H₂O = 1.37 mols O₂ x 18 mols H₂O / 25 mols O₂ x 18 g H₂O/mol = 17.8 g H₂O

Percent yield = actual yield / theoretical yield (x100%)

%yield = 8.68 g / 17.8 g (x100%) = 48.8% yield