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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?
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Solution
(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 NH3.
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
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