What should be the weight and moles of AgCl precipitate obtained on adding 500 ml of 0.20 M HCl in 30 g of ${\mathrm{AgNO}}_3\mathrm{solution}?\left({\mathrm{AgNO}}_3=170\right)$

$\begin{array}{l} {\mathrm{AgNO}}_3+\mathrm{HCl}\to \mathrm{AgCl}+{\mathrm{HNO}}_3\\ 1\mathrm{mole} + 1\mathrm{mole}\to 1\mathrm{mole}\\ \mathrm{given}, \frac{30}{170}\mathrm{moles}+\frac{500\times 0.2}{1000} \to ? \\ =0.176 + 0.1\mathrm{moles}\\ \mathrm{HCl} \mathrm{is} \mathrm{the} \mathrm{limiting} \mathrm{reagent}\\ \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{AgCl} \mathrm{formed} =0.1\times \mathrm{Molecular} \mathrm{wt}.=0.1\times 143.5=14.35\mathrm{g}\end{array}$