How many moles of AgCl would be obtained, when 100 mL of 0.1 M $\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{Cl}}_3$ is treated with excess of AgNO₃?

$\mathrm{CN}\text{ of }{\mathrm{Co}}^{3+}=6$

So, formulae of complex and reaction are as follows:

m moles of complex

$=100\mathrm{mL}\times 0.1\mathrm{M}=10\mathrm{m}\text{ moles. }$

$\therefore 1\mathrm{m}\text{ mole of complex }\Rightarrow 2\mathrm{m}\text{ moles of }\mathrm{AgCl}$

$\therefore 10\mathrm{m}\text{ mole of complex }\Rightarrow 20\mathrm{m}\text{ moles of }\mathrm{AgCl}$

$\Rightarrow 20\times {10}^{-3}=0.02\text{ moles }$