Question

How many moles of $AgCl$ would be obtained when $100$ mL of $0.1MCoC{l}_3.3N{H}_3$ is treated with excess $AgN{O}_3$?

Answer 1

Answer: 0

$AgN{O}_3+C{l}^{-}\to AgCl+N{O}_3^{-}$
Molarity $M=\frac{\text{weight}}{\text{Molecular weight}}\times \frac{1000}{V_{\left(mL\right)}}$
$\Rightarrow 0.1=n\times \frac{1000}{100}$
$\Rightarrow n=0.01$ moles