Đáp án:

\(\begin{array}{l}
{C_\% }MgC{l_2} = 3,04\% \\
{C_\% }CaC{l_2} = 18,648\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{P_1}:\\
MgC{l_2} + 2AgN{O_3} \to Mg{(N{O_3})_2} + 2AgCl(1)\\
CaC{l_2} + 2AgN{O_3} \to Ca{(N{O_3})_2} + 2AgCl(2)\\
{P_2}:\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
Mg{(OH)_2} \to MgO + {H_2}O\\
{n_{MgO}} = \dfrac{{0,32}}{{40}} = 0,008\,mol\\
{n_{MgC{l_2}}} = {n_{MgO}} = 0,008\,mol\\
{n_{AgCl}} = \dfrac{{14,35}}{{143,5}} = 0,1\,mol\\
{n_{AgCl(1)}} = 0,008 \times 2 = 0,016\,mol\\
{n_{AgCl(2)}} = 0,1 - 0,016 = 0,084\,mol\\
{n_{CaC{l_2}}} = \dfrac{{0,084}}{2} = 0,042\,mol\\
{C_\% }MgC{l_2} = \dfrac{{0,008 \times 2 \times 95}}{{50}} \times 100\%  = 3,04\% \\
{C_\% }CaC{l_2} = \dfrac{{0,042 \times 2 \times 111}}{{50}} \times 100\%  = 18,648\% 
\end{array}\)