In the reaction $2{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)+\mathrm{Ca}\left(\mathrm{OH}{)}_2\right(\mathrm{s})\to {\mathrm{CaCl}}_2(\mathrm{s})+2{\mathrm{NH}}_3(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ the mass of ${\mathrm{NH}}_3$ formed by heating a mixture containing 10.0 g each of ${\mathrm{NH}}_4\mathrm{Cl}$ and $\mathrm{Ca}(\mathrm{OH}{)}_2$ is (Given: Molar mass of $\left.\mathrm{Ca}=40\mathrm{g}{\mathrm{mol}}^{-1}\right)$

We have $\begin{array}{c}{\xi }_{max}\left({\mathrm{NH}}_4\mathrm{Cl}\right)=\frac{m_{{\mathrm{NH}}_4\mathrm{Cl}}/M_{{\mathrm{NH}}_4\mathrm{Cl}}}{v_{{\mathrm{NH}}_4\mathrm{Cl}}}=\frac{10.0\mathrm{g}/53.5\mathrm{g}{\mathrm{mol}}^{-1}}{2}=0.0935\mathrm{mol}\\ {\xi }_{max}\left(\mathrm{Ca}(\mathrm{OH}{)}_2\right)=\frac{m_{\mathrm{Ca}(\mathrm{OH}{)}_2}/M_{\mathrm{Ca}(\mathrm{OH}{)}_2}}{v_{\mathrm{Ca}(\mathrm{OH}{)}_2}}=\frac{10\mathrm{g}/74\mathrm{g}{\mathrm{mol}}^{-1}}{1}=0.1351\mathrm{mol}\end{array}$${\xi }_{max}\left({\mathrm{NH}}_4\mathrm{Cl}\right)<{\xi }_{max}\left(\mathrm{Ca}(\mathrm{OH}{)}_2\right)$, the limiting reagent is ${\mathrm{NH}}_4\mathrm{Cl}\text{. }$

$\begin{array}{r}\mathrm{\Delta }{n}_{{\mathrm{NH}}_3}=V_{{\mathrm{NH}}_3}{\xi }_{max}\left({\mathrm{NH}}_4\mathrm{Cl}\right)=2\times 0.0935\mathrm{mol}=0.1870\mathrm{mol}\\ m_{{\mathrm{NH}}_3}=\mathrm{\Delta }{n}_{{\mathrm{NH}}_3}{M}_{{\mathrm{NH}}_3}=\left(0.1870\mathrm{mol}\right)\left(17\mathrm{g}{\mathrm{mol}}^{-1}\right)=3.18\mathrm{g}\end{array}$