Qn. In haber process 30 litres 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?

N2 (g) +3 H2(g) →2 NH3
According to the Gay Lussac law , 1 L of the N2 gas would react with 3 L of H2 gas to produce 2 L of NH3.
Since 30 L of N2 and 30 L of H2 gas are present thus according to the balanced equation H2 is the limiting reagent .
Thus 30 L H2 gas on reaction with 10 L N2should produce 20 L NH3. But since it is given that only 50% of the expected product is obtained . This shows that only 15 L of H2 gas is reacting with 5 L N2 gas to give half of the expected product i.e. 10 LNH3 gas.

Thus composition of gases in the reaction mixture : 
Amount of N2 gas left in the reaction vessel = 30-5 = 25 L
Amount of H2 gas left in the reaction vessel = 30-15 = 15 L
Amount of NH3 gas produced in the reaction vessel = 10 L
 

Regards

Arun (askIITians forum expert)