Ques 14 plz

Dear Student,

The computation are expressed below,

$$\begin{gathered} LiCl.3N{H}_3\left(s\right)\iff LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right) \\ we have to compute for, \\ LiCl.N{H}_3\left(s\right)+2N{H}_3\left(g\right)\iff LiCl.3N{H}_3\left(s\right) \\ Initial moles 0.1 a 0 \\ Final moles 0 a-0.2 0.1 \\ at equn, {K_P}^{\text{'}}=\frac{1}{\left({P_{N{H}_3}}^{\text{'}}\right)}\Rightarrow \frac{1}{4}=\frac{1}{\left({P_{N{H}_3}}^{\text{'}}\right)} \\ {P^{\text{'}}}_{N{H}_3}=2 atm \\ PV=nRT \\ 2\times 5=n\times 0.0820\times 313 \\ n=0.3896\approx 0.40 \\ a-0.2=0.10 \\ a=0.30 \\ Initial moles of N{H}_3=0.30 \end{gathered}$$
0.728 mole answer is for Kp = 9 sq.atm and T=37 C so kindly check.

Regards.