Question

28 g of N₂ and 6g of H₂ were mixed. At equilibrium, 17g NH₃ was produced. The weights of N₂ and H₂ at equilibrium are respectively.

Answer 1

The reaction for the formation of ammonia is N₂​+3H₂ ​⇔ 2NH₃​.

1 mole (28 g) of nitrogen will react with 3 moles (6g) of hydrogen to form 2 moles (34g) of ammonia if the reaction is 100% completed.

But only 17 g of ammonia are actually formed.

Hence, the reaction is 50% complete.

Hence, 0.5 mole (14 g) of nitrogen and 3 moles (3g) of hydrogen will remain unreacted at equilibrium.

Comments

What is the answer if 24 g of N2 and 4 g of H2 is given in question in place of 28 and 6 respectively?