Forty per cent of a mixture of 0.2 mol of N₂ and 0.6 mol of H₂ react to give NH₃ according to the equation:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
at constant temperature and pressure. Then, the ratio of the

Question

Forty per cent of a mixture of 0.2 mol of N₂ and 0.6 mol of H₂ react to give NH₃ according to the equation:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
at constant temperature and pressure. Then, the ratio of the final volume to the initial volumes of gases is

Prashant Kotian · Accepted Answer

Correct answer: A. Initial total moles = 0.8 mol. Since they are in exact 1:3 stoichiometric ratio, a 40% reaction means 40% of the entire mixture converts. N₂ reacted = 0.08, H₂ reacted = 0.24, NH₃ formed = 0.16. Final moles = (0.2 - 0.08) + (0.6 - 0.24) + 0.16 = 0.12 + 0.36 + 0.16 = 0.64 mol. Ratio of volumes = Ratio of moles = 0.64 / 0.8 = 0.8 = 4/5. Therefore, correct answer is A.

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