Asked by qp
Consider the reaction.
mc015-1.jpg
How many grams of N2 are required to produce 100.0 L of NH3 at STP?
6.25 g
12.5 g
62.5 g
125.0 g
All Answers 1
Answered by
GPT-5 mini
AI
Reaction: N2 + 3H2 → 2NH3
At STP 1 mol gas = 22.4 L, so moles NH3 = 100.0 L / 22.4 L·mol−1 = 4.464 mol.
From stoichiometry, moles N2 = 4.464 × (1/2) = 2.232 mol.
Mass N2 = 2.232 mol × 28.02 g·mol−1 ≈ 62.5 g.
Answer: 62.5 g.
At STP 1 mol gas = 22.4 L, so moles NH3 = 100.0 L / 22.4 L·mol−1 = 4.464 mol.
From stoichiometry, moles N2 = 4.464 × (1/2) = 2.232 mol.
Mass N2 = 2.232 mol × 28.02 g·mol−1 ≈ 62.5 g.
Answer: 62.5 g.
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