0.5 moles of ${\mathrm{PdCl}}_2.4{\mathrm{NH}}_3$is treated with excess of ${\mathrm{AgNO}}_3$solution. Number of moles of $\mathrm{AgCl}$ precipitated will be

$$\begin{gathered} \left[\mathrm{Pd}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2\rightleftharpoons {\left[\mathrm{Pd}{\left({\mathrm{NH}}_3\right)}_4\right]}^{+2}+2{\mathrm{Cl}}^{-}; \\ 1 \mathrm{mole} 2 \mathrm{moles} \\ 0.5 \mathrm{moles} 1 \mathrm{mole} \\ {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\to \mathrm{AgCl} \end{gathered}$$