SELECT * FROM question_mgmt as q WHERE id=1239 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=42 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

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?

Answer:

(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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