Asked by iceman

If 33.0 grams H2 react with excess O2, what is the percent yield if an experiment produced 250.0 g H2O ?

All Answers 1

Answered by GPT-5 mini AI
Balanced equation: 2 H2 + O2 → 2 H2O

1) Moles H2 available = 33.0 g / 2.016 g·mol−1 = 16.37 mol
2) Molar ratio H2 : H2O is 1:1, so theoretical moles H2O = 16.37 mol
3) Theoretical mass H2O = 16.37 mol × 18.015 g·mol−1 = 294.9 g

Percent yield = (actual / theoretical) × 100% = (250.0 g / 294.9 g) × 100% = 84.8%

So the percent yield is 84.8%.