Calculate the mass of AgCl precipitated, when 25 ml of 35 % solution of AgNO 3 and 25 ml of 11 . 6 %…

${\mathrm{AgNO}}_{3 \left(\mathrm{aq}\right)}+{\mathrm{NaCl}}_{ \left(\mathrm{aq}\right)}\to {\mathrm{AgCl}}_{ \left(\mathrm{s}\right)}+{\mathrm{NaNO}}_{3 \left(\mathrm{aq}\right)}$

The above equation is balanced, and we can see that when one mole of ${\mathrm{AgNO}}_3$ and one mole of $\mathrm{NaCl}$ react then one mole of $\mathrm{AgCl}$ is precipitated.

So, the reactant which is present in lesser quantity determines the amount of $\mathrm{AgCl}$ formed.

$25 \mathrm{ml}$ of $35\% {\mathrm{AgNO}}_3$

$=25\times \frac{35}{100}=8.75 \mathrm{g}$ of ${\mathrm{AgNO}}_3$

$25 \mathrm{ml}$ of $11.6\% \mathrm{NaCl}$

$=25\times \frac{11.6}{100}=2.9 \mathrm{g}$ of $\mathrm{NaCl}$

As $1$ mole $\mathrm{NaCl}$ $=58.5 \mathrm{g}$

So, $2.9 \mathrm{g}$ of $\mathrm{NaCl}$ $=\frac{2.9}{58.5}=0.049$ moles

Therefore, $0.049$ moles of $\mathrm{AgCl}$ is precipitated.

Hence, amount of $\mathrm{AgCl}$ $$\begin{gathered} =\mathrm{number} \mathrm{of} \mathrm{moles} \times \mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{AgCl} \\ =0.049\times 143 \mathrm{g} \\ =7 \mathrm{g} \end{gathered}$$