How many moles of AgCl would be obtained, when 100 mL of 0.1 M CoNH35Cl3 is treated with excess of AgNO3?

CN of Co3+=6So, formulae of complex and reaction are as follows:    m moles of complex=100mL×0.1M=10m moles. ∴1m mole of complex ⇒2m moles of AgCl∴10m mole of complex ⇒20m moles of AgCl⇒20×10−3=0.02 moles