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 $$NH_{ 3 }produced\; if \; \times \;10^{ 3 }$$ g N2 reacts with $$1 \; \times \;10^{ 3 }g \;of\; H_2?$$

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass.

Asked by Abhisek | 1 year ago |  90

Solution :-

(i) Balance the given equation:

$$N_{ 2 }\;(g) \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g)$$

Thus, 1 mole (28 g) of N2 reacts with 3 mole (6 g) of H2 to give 2 mole (34 g) of $$NH_{ 3 }$$

$$2 \; \times \;10^{ 3 } g\; of N_2 \;will \;react \;with$$

$$\dfrac{ 6}{ 28 } \; \times \; 2 \; \times \; 10^{ 3 } g NH_3$$

$$2 \; \times \;10^{ 3 }g$$  of N2 will react with 428.6 g of H2.

Given:

Amt of H2 = $$1 \; \times \;10^{ 3 }$$

28 g of $$N_{ 2 }$$ produces 34 g of $$NH_{ 3 }$$

Therefore, mass of$$NH_{ 3 }$$  produced by 2000 g of $$NH_{ 2 }$$

$$\dfrac{ 34 \; g }{ 28 \; g } \; \times \; 2000g$$

2430 g of $$NH_{ 3 }$$

(ii) $$H_{ 2 }$$ is the excess reagent.

Therefore, $$H_{ 2 }$$will not react.

(iii) Mass of H2 unreacted

$$1 \; \times \;10^{ 3 }– 428.6 g$$

= 571.4 g

Answered by Pragya Singh | 1 year ago

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