Question

What mass of Silver chloride will be obtained by adding an excess of Hydrochloric acid to a solution of 0.34g of Silver nitrate? ($\mathrm{CI}=35.5,\mathrm{Ag}=108,\mathrm{N}=14,\mathrm{O}=16,\mathrm{H}=1)$

Answer 1

Step 1

$$\begin{gathered} {\mathrm{AgNO}}_3\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)\to \mathrm{AgCl}\left(\mathrm{s}\right)+\mathrm{HCl}\left(\mathrm{aq}\right) \\ \mathrm{Silver}\mathrm{nitrate}\mathrm{Hydrochloric}\mathrm{acid}\mathrm{Silver}\mathrm{chloride}\mathrm{Hydrchloric}\mathrm{acid} \end{gathered}$$
$$\begin{gathered} (108+14+48)\mathrm{g}(108+35.5)\mathrm{g} \\ 170\mathrm{g}143.5\mathrm{g} \end{gathered}$$

Step 2:

Find the mass:
170 g of ${\mathrm{AgNO}}_3$ produces 143.5 g $\mathrm{AgCl}$
Therefore, 0.34 g of ${\mathrm{AgNO}}_3$ will produce $\mathrm{AgCl}$ = $\frac{143.5x0.34}{170}=0.287\mathrm{g}$

Hence, 0.287 g silver chloride will be obtained.