Question

[1] calculate the volume of NH3 gas produced at STP when 140 g of N2 reacts with 30 g of H2?

Answer 1

Moles of N₂ = 140 / 28 = 5 mol
Moles of H₂ = 30 / 2 = 15 mol

Balanced equation of formation of NH₃
N₂ + 3H₂ → 2NH₃
From above balanced equation we can say that for formation of two mole of NH₃ 1 mol of N₂ and 3 mol of H₂ is required.
We have 5 and 15 mol of N₂ and H₂ respectively.
There is no limiting reagent.

Now,
3 mol of H₂ gives 2 moles of NH₃
15 mol of H₂ will give 10 moles of NH₃

Moles of NH₃ produced is 10

At STP
One mole of any gas occupies 22.4 Litre
10 mole of NH₃ at STP will occupy 224 Litre

Volume of NH₃ gas formed is 224 Litres