Question

56 kg of N₂(g) and 10 kg of H₂(g) are mixed to produce NH₃(g).Calculate the number of moles of ammonia gas formed. 

(Atomic mass/g mol^–1 N = 14, H = 1)

Answer 1

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Since, 28 kg N₂ reacts with 6 kg of H₂

Therefore, 56 kg N₂ reacts with \(\frac{6}{28}\) x 56 = 12 kg of H₂

But we have only 10 kg of H₂, therefore, H₂ is limiting reactant.

Also, 6 kg of H₂ will give 2 moles of NH₃ 

Hence, 10 kg of H₂ will give \(\frac{2}{6}\) × 10 

= 3.33 moles