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)

(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Answer 1

(i) Balancing the given chemical equation,

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 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 NH_3.

Hence, mass of ammonia produced by 2000 g of N₂

= 2428.57 g

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

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

= 571.4 g