Question 24

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?


(i) Balancing the given chemical equation,



Total mass of Ammonia = 2((14) +3(1))= 34 g

From the chemical equation,we can write

28gm of N2 reacts with 6gm of H2 to produce ammonia= 34g

Or  1 gm of  N2 reacts with 1gm of H2 to produce ammonia= 34/28*1

Or  when 2.00x103 g of  N2 reacts with 1.00x103 gm of H2 to produce ammonia


=34/28 *2.00x103=2428.57g


hence 2.00 × 103 g of dinitrogen will react with  1.00x103 g of dihydrogen to give 2428.57 g of ammonia

Given,  Amount of dihydrogen = 1.00 × 103 g

Hence, N2 is the limiting reagent.


(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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