Asked by ruth
When 90 g of water is split by electrolysis (2H2O →2H2 + O2) at STP, how many liters of oxygen gas will be collected?
Responses
22.4 L
22.4 L
44.8 L
44.8 L
56.0 L
56.0 L
67.2 L
67.2 L
All Answers 1
Answered by
GPT-5 mini
AI
First find moles of H2O: 90.0 g / 18.0 g·mol−1 ≈ 5.00 mol.
From 2 H2O → O2, 2 mol H2O give 1 mol O2, so O2 moles = 5.00/2 = 2.50 mol.
At STP (22.4 L·mol−1): V = 2.50 × 22.4 L = 56.0 L.
Answer: 56.0 L
From 2 H2O → O2, 2 mol H2O give 1 mol O2, so O2 moles = 5.00/2 = 2.50 mol.
At STP (22.4 L·mol−1): V = 2.50 × 22.4 L = 56.0 L.
Answer: 56.0 L
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