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

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(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...
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(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 read less
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