Question

What weight of $\mathrm{AgCl}$ will be precipitated when a solution containing $4.77\mathrm{g}$ of $\mathrm{NaCl}$ is added to a solution of $5.77\mathrm{g}$ of ${\mathrm{AgNO}}_3$?

Answer 1

Step 1: Write the reactions and determine the molar masses of all compounds

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

Molar masses:

$$\begin{gathered} \mathrm{AgCl}=107.9\mathrm{u}+35.5\mathrm{u}=143.4\mathrm{u} \\ \mathrm{NaCl}=23\mathrm{u}+35.5\mathrm{u}=58.5\mathrm{u} \\ {\mathrm{AgNO}}_3=107.9\mathrm{u}+14\mathrm{u}+3\times 16\mathrm{u}=169.9\mathrm{u} \end{gathered}$$

Step 2: Calculate the number of moles of ${\mathrm{AgNO}}_3$ and $\mathrm{NaCl}$

Number of moles $=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}$

Number of moles of $\mathrm{NaCl}$ $=\frac{4.77}{58.4}=0.08\mathrm{moles}$

Number of moles of ${\mathrm{AgNO}}_3$$=\frac{5.77}{169.9}=0.034\mathrm{moles}$

Step 3: Calculate the number of moles of $\mathrm{AgCl}$

According to the law of conservation of mass

Number of moles of $\mathrm{AgCl}$$=\frac{\mathrm{mol}{\mathrm{AgNO}}_3\times \mathrm{molar}\mathrm{ratio}\mathrm{AgCl}}{\mathrm{molar}\mathrm{ratio}\mathrm{NaCl}}$

Number of moles of $\mathrm{AgCl}$$=\frac{0.034\mathrm{moles}\times 1}{1}=0.034\mathrm{moles}$

Step 4: Calculate the mass of $\mathrm{AgCl}$

Mass $=\mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{molar}\mathrm{mass}$

Mass $=0.034\mathrm{moles}\times 143.4\mathrm{g}/\mathrm{mol}=4.87\mathrm{g}$

Hence, the weight of $\mathrm{AgCl}$ will be precipitated when a solution containing $4.77\mathrm{g}$ of $\mathrm{NaCl}$ is added to a solution of $5.77\mathrm{g}$ of ${\mathrm{AgNO}}_3$ is $4.87\mathrm{g}$.