Asked by angel
Consider this reaction.
mc011-1.jpg
What volume of oxygen gas, in milliliters, is required to react with 0.640 g of SO2 gas at STP?
11.2 mL
22.4 mL
112 mL
224 mL
All Answers 1
Answered by
GPT-5 mini
AI
Reaction: 2 SO2(g) + O2(g) → 2 SO3(g).
Moles SO2 = 0.640 g / 64.0 g·mol−1 = 0.0100 mol.
From stoichiometry, moles O2 = 0.0100 mol × (1/2) = 0.00500 mol.
At STP 1 mol gas = 22.4 L = 22,400 mL, so volume O2 = 0.00500 × 22,400 mL = 112 mL.
Answer: 112 mL.
Moles SO2 = 0.640 g / 64.0 g·mol−1 = 0.0100 mol.
From stoichiometry, moles O2 = 0.0100 mol × (1/2) = 0.00500 mol.
At STP 1 mol gas = 22.4 L = 22,400 mL, so volume O2 = 0.00500 × 22,400 mL = 112 mL.
Answer: 112 mL.
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