Asked by qp
Consider the equation for the formation of water.
2H2 + O2 Right arrow. 2H2O
What is the theoretical yield of H2O if 130 g of H2O is produced from 18 g of H2 and an excess of O2?
18 g
81 g
130 g
160 g
All Answers 1
Answered by
GPT-5 mini
AI
2 H2 → 2 H2O so 1 mol H2 → 1 mol H2O.
Moles H2 = 18 g ÷ 2.016 g·mol−1 ≈ 8.93 mol → moles H2O = 8.93 mol.
Mass H2O = 8.93 mol × 18.016 g·mol−1 ≈ 160.8 g ≈ 160 g.
Answer: 160 g. (Actual yield 130 g gives percent yield ≈ 130/160 = 81%.)
Moles H2 = 18 g ÷ 2.016 g·mol−1 ≈ 8.93 mol → moles H2O = 8.93 mol.
Mass H2O = 8.93 mol × 18.016 g·mol−1 ≈ 160.8 g ≈ 160 g.
Answer: 160 g. (Actual yield 130 g gives percent yield ≈ 130/160 = 81%.)
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