When $50\text{ mL}$ of a $16.9\%\text{ (w/v)}$ solution of $\text{AgNO}_3$ is mixed with $50\text{ mL}$ of $5.8\%\text{ (w/v) NaCl}$ solution, then the mass of precipitate formed is: ($\text{Ag} = 107.8$, $\text{N} = 14$, $\text{O} = 16$, $\text{Na} = 23$, $\text{Cl} = 35.5$)

Mass of $\text{AgNO}_3$ in $50\text{ mL}$ of $16.9\%\text{ (w/v)}$ solution = $\frac{16.9}{100} \times 50 = 8.45\text{ g}$. Molar mass of $\text{AgNO}_3 = 107.8 + 14 + (16 \times 3) = 169.8\text{ g/mol}$. Moles of $\text{AgNO}_3 = \frac{8.45}{169.8} \approx 0.0497\text{ mol}$.

Mass of $\text{NaCl}$ in $50\text{ mL}$ of $5.8\%\text{ (w/v)}$ solution = $\frac{5.8}{100} \times 50 = 2.9\text{ g}$. Molar mass of $\text{NaCl} = 23 + 35.5 = 58.5\text{ g/mol}$. Moles of $\text{NaCl} = \frac{2.9}{58.5} \approx 0.0495\text{ mol}$.

The balanced chemical equation is: $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl}\downarrow + \text{NaNO}_3$

From the stoichiometry, $1\text{ mole}$ of $\text{AgNO}_3$ reacts with $1\text{ mole}$ of $\text{NaCl}$. Here, the moles of $\text{NaCl}$ ($0.0495\text{ mol}$) are slightly less than the moles of $\text{AgNO}_3$ ($0.0497\text{ mol}$), making $\text{NaCl}$ the limiting reagent.

Moles of $\text{AgCl}$ precipitate formed = Moles of $\text{NaCl}$ reacted = $0.0495\text{ mol}$. Molar mass of $\text{AgCl} = 107.8 + 35.5 = 143.3\text{ g/mol}$. Mass of $\text{AgCl}$ formed = Moles $\times$ Molar mass = $0.0495\text{ mol} \times 143.3\text{ g/mol} \approx 7.09\text{ g} \approx 7\text{ g}$.