# Question 05592

Aug 27, 2016

Now we can proceed framing ICE table

${N}_{2} \left(g\right) \text{ "+" "3H_2(g)" "rightleftharpoons" } 2 N {H}_{3} \left(g\right)$

I $\text{ "0.2" mol " 0.6" mol " 0" mol}$

C $\text{ "-0.2⋅40%"mol " -0.6⋅40%"mol "+0.16"mol}$

E $\text{ "0.12" mol " 0.36" mol "0.16"mol}$

As 40% reactants react

N_2→"reacted "0.2*40%=0.08" mol,Leaving 0.12 mol"

H_2→"reacted "0.6*40%=0.24" mol, Leaving 0.36 mol"

NH_3→"produced 0.16 mol"# ( twice the no.of moles of nitrogen reacted.)

Initial total number of moles = 0.8mol

Final total number of moles =$0.12 + 0.36 + 0.16 = 0.64 \text{ mol}$

As the temperature and pressure remain unaltered the ratio of final and initial volumes of gases will be equal to the ratio of their respective total no. of moles.So the ratio becomes

$\textcolor{red}{0.64 : 0.80 = 4 : 5}$