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