Question

Consider the chemical equation.


2H2 + O2 Right arrow. 2H2O


What is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2?
Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
56.5%
59.0%
88.5%
99.7%

Answer 1

1) Moles: H2 = 11.0 g / 2.016 g·mol⁻¹ = 5.456 mol. O2 = 95.0 g / 32.00 g·mol⁻¹ = 2.969 mol.

2) Stoichiometry: 2 H2 → 2 H2O so 1 mol H2 → 1 mol H2O. H2 is limiting (needs for O2 would be 2×2.969 = 5.938 mol > 5.456 mol).

3) Theoretical H2O = 5.456 mol × 18.016 g·mol⁻¹ = 98.30 g.

4) Percent yield = (87.0 g / 98.30 g) × 100% = 88.5%.

Answer: 88.5%.