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N2(g) + H2(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?
Asked by Arati Chaudhari Last Modified
Answer 
Nurullmobeen
(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 NH3.
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
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