Question

Question 1.24 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?

Class XI Some Basic Concepts of Chemistry Page 24

Answer 1

$N_2(g)+3H_2(g)----->2NH_3(g) \\1mol \: \: \: \: \: \: \: \: 3mol \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2mol \\ 28g \: \: \: \: \: \: \: \: \: \: \: 6g \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 34g$

here , we observed that,
28 g of N2 reacts with 6g of H2 .
∴ 1g of N2 reacts with 6/28 g of H2
∴ 2000g of N2 will react with 2000× 6/28 = 428.57 of H2


but H2 is given 1 × 10³g > 428.57 g
e.g., H2 is excess available .
hence, N2 is the limiting reagent. it means N2 limits the production of amount of anomia .

28g of N2 produces 34 g of NH3
1g of N2 produces 34/28 g of NH3
2000g of N2 will be produce 34/28 × 2000 = 2428.57 g of NH3 .


(ii) H2 is in excess so, it remains unreacted .
(iii) amount of H2 unreacted = 1000 - 428.57
= 571.43 g

Answer 2

Answer:

Explanation:

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