Question

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

\[\ce{N2 (g) + 3H2 (g) → 2NH3 (g)}\]

(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,

\[\ce{N2 (g) + 3H2 (g) → 2NH3 (g)}\]

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 `(6  "g")/(28  "g")xx2.00xx10^3  "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 NH_3.

Hence, mass of ammonia produced by 2000 g of N₂ `(34  "g")/(28  "g")xx2000  "g"`

= 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