Question

15.In Haber’s process if 0.06 mole of hydrogen disappears in 10 min the no. Of moles of ammonia formed in 0.3 min is a. 0.018 b. 0.0012 c. 0.04 d. 0.036

Answer 1

Answer:

b.0.0012

Explanation:

haber's process : N2+3H2--->2NH3

d(H2) = change in moles of H2

d(NH3) = change in moles of ammonia

dt = change in time

-1/3[d(h2)/dt]=1/2[d(nh3)/dt]

-1/3[-0.06/10]=1/2[d(nh3)/0.3]

by solving we get

d(NH3) = 0.0012

hope it is helpful