What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO₃ is mixed with 50 mL of 5.8% NaCl solution?
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

$\begin{array}{l}{\mathrm{n}}_{{\mathrm{AgNO}}_3}=\frac{50\times 16.9}{100\times 169.8}\approx 0.05\text{ Mole }\\ {\mathrm{n}}_{\mathrm{NaCl}}=\frac{50\times 5.8}{100\times 58.5}\approx 0.05\text{ Mole }\\ \therefore {\mathrm{AgNO}}_3+\mathrm{NaCl}⟶\mathrm{AgCl}+{\mathrm{NaNO}}_3\end{array}$
$\begin{array}{l}\mathrm{t}=0;\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.05\text{ mole }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.05\text{ mole }\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ \mathrm{t}=\mathrm{t};\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0.05\text{ mole }\end{array}$
$\begin{array}{l}\therefore \text{ Mass of }\mathrm{AgCl}=0.05\times 143.3\\ =7.16\\ \approx 7\mathrm{g}\end{array}$