Asked by qp
Consider this reaction.
mc030-1.jpg
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
1860 mL
2880 mL
3710 mL
45,100 mL
All Answers 1
Answered by
GPT-5 mini
AI
Balanced equation: 2 Mg + O2 → 2 MgO.
Moles Mg = 4.03 g / 24.305 g·mol⁻¹ = 0.1658 mol.
Moles O2 required = 0.1658 mol Mg × (1 mol O2 / 2 mol Mg) = 0.0829 mol O2.
At STP 1 mol gas = 22.4 L, so V = 0.0829 mol × 22.4 L·mol⁻¹ = 1.857 L = 1857 mL ≈ 1860 mL.
Answer: 1860 mL.
Moles Mg = 4.03 g / 24.305 g·mol⁻¹ = 0.1658 mol.
Moles O2 required = 0.1658 mol Mg × (1 mol O2 / 2 mol Mg) = 0.0829 mol O2.
At STP 1 mol gas = 22.4 L, so V = 0.0829 mol × 22.4 L·mol⁻¹ = 1.857 L = 1857 mL ≈ 1860 mL.
Answer: 1860 mL.
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