Asked by ruth
You have 39 g of sodium azide in an air bag[2NaN3 (s) →2Na (s) + 3N2(g)].
How many liters of nitrogen will it make at STP?
Responses
13.4 L
13.4 L
8.9 L
8.9 L
20.2 L
20.2 L
22.4 L
22.4 L
All Answers 1
Answered by
GPT-5 mini
AI
2 NaN3 → 2 Na + 3 N2
Molar mass NaN3 = 65.0 g·mol⁻¹, so moles NaN3 = 39 g / 65.0 g·mol⁻¹ = 0.60 mol.
From stoichiometry: 1 mol NaN3 → 1.5 mol N2, so N2 = 0.60 × 1.5 = 0.90 mol.
At STP (22.4 L·mol⁻¹): V = 0.90 × 22.4 = 20.16 L ≈ 20.2 L.
Answer: 20.2 L
Molar mass NaN3 = 65.0 g·mol⁻¹, so moles NaN3 = 39 g / 65.0 g·mol⁻¹ = 0.60 mol.
From stoichiometry: 1 mol NaN3 → 1.5 mol N2, so N2 = 0.60 × 1.5 = 0.90 mol.
At STP (22.4 L·mol⁻¹): V = 0.90 × 22.4 = 20.16 L ≈ 20.2 L.
Answer: 20.2 L
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