Question

in haber process 30 litres of di hydrogen and 30 litres of di nitrogen were taken for reaction which yielded only 50% of the expected product. what will be the composition of the gaseous mixture under the aforesaid condition in the end

Answer 1

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Answer 2

3H₂ + N₂ = 2NH₃

When there is 100% yield,

1 mol of N2 reacts with 3 moles of H₂. Hence, H₂ will be the limiting reagent.

Then,

3H₂ will give = 2 moles of NH₃

30L of H₂ will give = 2/3 × 30

= 20 L NH₃

However, when there is 50% yield,

NH₃ = 1/2 × 20

= 10L NH₃

H₂ = 1/2 × 30

= 15 L H₂

N2 = 20 + 1/2 × 10

= 25 L N₂

Hence, 10 L NH₃, 15 L H₂ and 25 L N₂ will be formed.