### Q.24:- 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 2.00×10^{3}g dinitrogen reacts with 1.00×10^{3}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 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).

∴ 2000 g of N

_{2 }will react with_{ }H_{2 }= 6/28 ×200g = 428.6g. Thus, here N_{2 }is the limiting reagent while_{ }H_{2 }is in excess.28g of N

_{2}produce 34g of NH_{3.}∴2000g of N

_{2}will produce = 34/28×2000g = 2428.57 g of NH_{3.}(ii) N

_{2}is the limiting reagent and H_{2}is the excess reagent. Hence, H_{2}will remain unreacted.(iii) Mass of dihydrogen left unreacted = 1000g – 428.6g = 571.4 g