Question

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

Answer 1

Step 1: Given information

• Volume of $16.9\%(\mathrm{w}/\mathrm{v})$ ${\mathrm{AgNO}}_3$ (silver nitrate) solution is $50\mathrm{mL}$.
• Volume of $5.8\%(\mathrm{w}/\mathrm{v})\mathrm{NaCl}$(sodium chloride) solution is $50\mathrm{mL}$.

Step 2: Calculate the number of moles of silver nitrate

• The molar mass of ${\mathrm{AgNO}}_3$ is $169.8\mathrm{g}{\mathrm{mol}}^{-1}$.
• The number of moles of silver nitrate is as follows:

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{moles}& =& \frac{\mathrm{Volume}\times \mathrm{Weight}\mathrm{by}\mathrm{volume}\mathrm{percentage}}{\mathrm{Molar}\mathrm{mass}}\\ & =& \frac{50\bcancel{\mathrm{mL}}\times 16.9\bcancel{\mathrm{g}}}{100\bcancel{\mathrm{mL}}\times 169.8\bcancel{\mathrm{g}}{\mathrm{mol}}^{-1}}\\ & =& 0.0497\mathrm{mol}\end{array}$

Step 3: Calculate the number of moles of Sodium chloride

• The molar mass of $\mathrm{NaCl}$ is $58.5\mathrm{g}{\mathrm{mol}}^{-1}$.
• The number of moles of sodium chloride is as follows:

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{moles}& =& \frac{\mathrm{Volume}\times \mathrm{Weight}\mathrm{by}\mathrm{volume}\mathrm{percentage}}{\mathrm{Molar}\mathrm{mass}}\\ & =& \frac{50\bcancel{\mathrm{mL}}\times 5.8\bcancel{\mathrm{g}}}{100\bcancel{\mathrm{mL}}\times 58.5\bcancel{\mathrm{g}}{\mathrm{mol}}^{-1}}\\ & =& 0.0496\mathrm{mol}\end{array}$

Step 4: Calculate the mass of precipitate

• The reaction is as follows:

${\mathrm{AgNO}}_3+\mathrm{NaCl}\to \mathrm{AgCl}+{\mathrm{NaNO}}_3$

• From the reaction $1\mathrm{mol}$ of ${\mathrm{AgNO}}_3$ precipitate out $1\mathrm{mol}$ of $\mathrm{AgCl}$.
• Therefore, $0.0497\mathrm{mol}$ of ${\mathrm{AgNO}}_3$ precipitate out $0.0497\mathrm{mol}$ of $\mathrm{AgCl}$.
• The molar mass of $\mathrm{AgCl}$ (silver chloride) is $143.3\mathrm{g}{\mathrm{mol}}^{-1}$.
• The mass of $\mathrm{AgCl}$ is as follows:

$\begin{array}{rcl}\mathrm{Mass}& =& \mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{Molar}\mathrm{mass}\\ & =& 0.0497\bcancel{\mathrm{mol}}\times 143.3\mathrm{g}\bcancel{{\mathrm{mol}}^{-1}}\\ & =& 7.12\mathrm{g}\end{array}$

Therefore the mass of precipitate $\mathbf{AgCl}$ is $\mathbf{7}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{g}$.