Asked by qp
Consider the equation.
mc016-1.jpg
Which represents the correct mass-volume relationship at STP?
32.00 g of O2 react with an excess of H2 to produce (2 ´ 22.4) L of H2O.
2.00 g of H2 react with an excess of O2 to produce (2 ´ 22.4) L of H2O.
32.00 g of O2 react with an excess of H2 to produce 22.4 L of H2O.
(2 ´ 2.00) g of H2 react with an excess of O2 to produce 22.4 L of H2O.
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GPT-5 mini
AI
Answer: 1) 32.00 g of O2 react with an excess of H2 to produce (2 × 22.4) L of H2O.
Reason: 32.00 g O2 = 1.00 mol O2. From 2H2 + O2 → 2H2O, 1 mol O2 gives 2 mol H2O. At STP 1 mol gas = 22.4 L, so 2 mol H2O = 2 × 22.4 = 44.8 L. The other choices give wrong mole/volume relationships.
Reason: 32.00 g O2 = 1.00 mol O2. From 2H2 + O2 → 2H2O, 1 mol O2 gives 2 mol H2O. At STP 1 mol gas = 22.4 L, so 2 mol H2O = 2 × 22.4 = 44.8 L. The other choices give wrong mole/volume relationships.
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