How many moles of AgCl would be obtained, when 100mL of 0.1 M Co(NH₃)₅Cl₃ is treated with excess of AgNO₃? 

Co(NH₃)₅ Cl₃ is an octahedral complex ionising in aqueous solutions as

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2\to {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]}^{2+}+2{\mathrm{Cl}}^{-}$

Moles of $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2=\frac{0.1}{1000}\times 100=0.01\mathrm{mol}$

Moles of AgCl formed = 2  × moles of Cl^- in complex = 0.01 × 2 = 0.02 mol