Dinitrogen and di

The reaction given in the question is not balanced.

Balancing the above chemical reaction:

N₂ + 3H₂→ 2NH₃

(i) Finding mass of NH₃ produced:

Atomic Mass of Hydrogen = 1g

Atomic Mass of Nitrogen = 14g

Molecular Mass of NH₃ = 14 + 3

= 17 g

     N₂    +     3H₂→ 2NH₃
     28gm      6gm       34gm

From the chemical equation we observe that 28g of nitrogen react with 6g of hydrogen to give 34g of ammonia

Therefore 1 mole of dinitrogen reacts with 3 moles dihydrogen to give 2 moles of ammonia.

Therefore, amount of dihydrogen required to react with 2000g of dinitrogen = [6×2000]/28

= 12000/28

= 428.57g

Therefore, 2000g of dinitrogen will produce = [34×2000]/28

= 68000/28

= 2428.57g of Ammonia

Therefore 2428.57g of ammonia is produce when 2000g of dinitrogen is used.

(ii) Since dinitrogen is present in lesser amount(428.57 gm), it is the limiting reagent and dihydrogen is the excess reagent and hence H₂ will be left unreacted.

(iii) Mass of dihydrogen left unreacted = 1000 – 428.57

= 571.43g