Question

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: \(N_2 (g) + H_2 (g) → 2NH_3 (g)\)

1. Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.
2. Will any of the two reactants remain unreacted?
3. If yes, which one and what would be its mass?

Answer 1

(i) Balancing the given chemical equation,
\(N_2(g) + 3H_2(g) → 2NH_3(g)\)
From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.
⇒ 2.00 × 10³ g of dinitrogen will react with \(\frac {6\  g}{28\  g} × 2.00 × 10^3 g\) dihydrogen i.e.,
2.00 × 10³ g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 10³ g
Hence, N₂ is the limiting reagent.
∴ 28 g of N₂ produces 34 g of NH₃.
Hence, mass of ammonia produced by 2000 g of N₂ = \(\frac {34\  g}{28\ g} × 2000\  g = 2428.57\  g\)

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(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.

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(iii) Mass of dihydrogen left unreacted = 1.00 × 10³ g - 428.6 g = 571.4 g