Question

The electrolysis of water forms H2 and O2.


2H2O Right arrow. 2H2 + O2


What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
15.1%
33.8%
60.1%
67.6%

Answer 1

Step 1: Moles H2O = 17.0 g / 18.015 g·mol⁻¹ = 0.944 mol (approx).

Step 2: From 2 H2O → O2, 2 mol H2O → 1 mol O2, so moles O2 theoretical = 0.944 / 2 = 0.472 mol.

Step 3: Mass O2 theoretical = 0.472 mol × 32.00 g·mol⁻¹ = 15.1 g.

Percent yield = (actual / theoretical) × 100 = (10.2 / 15.1) × 100 = 67.6%.

Answer: 67.6%