What is the mass of the precipitate formed when 50 mL of 16.9 %  solution of ${\mathrm{AgNO}}_3$ is mixed with 50 mL of 5.8% NaCl solution? 

$(\mathrm{Ag}=107.8, \mathrm{N}=14,\mathrm{O}=16,\mathrm{Na}=23,\mathrm{Cl}=35.5)$

$\underset{50\mathrm{ml}\text{ of16.9\% }}{{\mathrm{AgNO}}_3}+\underset{50\mathrm{mlofl}5.8\mathrm{\%}}{\mathrm{NaCl}}\to \underset{\text{ (ppt) }}{\mathrm{AgCl}}+{\mathrm{NaNO}}_3$
Number of moles of
${\mathrm{AgNO}}_3=\frac{\mathit{50}\mathit{\times }\mathit{16}\mathit{.}\mathit{9}}{\mathit{100}\mathit{\times }\mathit{170}}\mathit{=}\mathit{0}\mathit{.}\mathit{049}\mathit{\approx }\mathit{0}\mathit{.}\mathit{05}$
Number of moles of $\mathrm{NaCl}=\frac{50\times 5.8}{100\times 58.5}=0.0495\approx 0.05$
Number of moles of AgCl precipitates
= 0.05 moles
=0.05 x 143.5 = 7.175 g $\approx$ 7 g