In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?

Options

A. 20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
B. 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
C. 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
D. 20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen

Answer

C

Explanation

`N_2 (g) +3H_(2) (g) hArr 2NH_3 (g)`
`30 - x" " 30 - 3x " " 2x` at equilibrium 1 mole of `N_2(g)` reacts with 3 moles of `H_2(g)` and produce 2 moles of `NH_3(g)` . Since only 30 litres of `H_2(g)` is present so `H_2(g)` is the limiting reagent. Thus30 litres of `H_2(g)` will react with 10 litres of `N_2(g)` and produce 20 litres of `NH_3(g)` . But since it is given that only 50% of the expected product obtained. So
`2x = 10 implies x = 5`
Volume of Ammonia = 10
Volume of Nitrogen = 30 - 5 = 25
Volume of Hydrogen = 30 - 15 = 15 .