Question

During preparation of ammonia by haber's process 30L of hydrogen gas and 30L of nitrogen gas are mixed.The yield of ammonia was 50%.Find the composition of gaseous mixture

Answer 1

The reaction for the process is
N2 + 3H2 = 2NH3

i.e. 1 of N2 mole combines with 3 moles of H2 to form 2 moles of NH3

I.e. 22.4L of N2 combines with 67.2L of H2 to form 44.8L of NH3

Thus 30L of N2 will require 67.2*30/22.4 = 90L of H2, but we have only 30L of H2 hence H2 is the limiting reagent.

no of moles of H2 = 30/22.4 = 1.34 moles

3 moles of H2 gives 2 moles of NH3
thus, 1.34 moles of H2 gives 1.34*2)/3 = 0.89 moles

Amount of NH3 produced = 0.89*17 = 15.18 * 50%. = 7. 59g or 10L

Composition of gases :

N2 = H2 = 1.34/(1.34+1.34+0.89) = 37.5%

NH3 = 25%