Question

1. for the reaction: n₂(g) + 3 h₂(g) → 2nh₃(g)

a) if 2.5 liters of n₂ react with sufficient h₂, how many liters of nh₃ will form (p and t remain constant)?
b) if the volume of n₂ given was measured at stp, how many grams of nh₃ would be produced?

Explanation:

Step1: Relate N₂ and NH₃ volumes

Mole ratio N₂:NH₃ = 1:2. At constant P, T, volume ∝ moles.
Volume of NH₃ = 2×Volume of N₂ = 2×2.5 L = 5 L.

Step2: Find moles of N₂ at STP

1 mol gas = 22.4 L at STP.
Moles of N₂ = $\frac{2.5\ \text{L}}{22.4\ \text{L/mol}} ≈ 0.1116\ \text{mol}$.

Step3: Calculate moles of NH₃

Moles of NH₃ = 2×moles of N₂ ≈ 2×0.1116 mol = 0.2232 mol.

Step4: Compute mass of NH₃

Molar mass of NH₃ = 14 + 3×1 = 17 g/mol.
Mass = moles×molar mass ≈ 0.2232 mol×17 g/mol ≈ 3.79 g.

Answer:

a) 5 liters
b) Approximately 3.8 grams (or 3.79 grams)