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: Find mole ratio of NH₃ to N₂

From reaction: $\ce{N2 + 3H2 -> 2NH3}$, ratio $\frac{V_{NH3}}{V_{N2}} = \frac{2}{1}$

Step2: Calculate volume of NH₃

$V_{NH3} = 2.5\,\text{L} \times 2 = 5\,\text{L}$

Step3: Moles of N₂ at STP

Molar volume = 22.4 L/mol, $n_{N2} = \frac{2.5\,\text{L}}{22.4\,\text{L/mol}} \approx 0.1116\,\text{mol}$

Step4: Moles of NH₃ produced

$n_{NH3} = 2 \times n_{N2} \approx 2 \times 0.1116 = 0.2232\,\text{mol}$

Step5: Mass of NH₃

Molar mass of NH₃ = 17 g/mol, $mass = 0.2232\,\text{mol} \times 17\,\text{g/mol} \approx 3.79\,\text{g}$

Answer:

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