Đáp án:
\(\begin{array}{l}
{C_\% }MgC{l_2} = 9,5\% \\
{C_\% }CuC{l_2} = 13,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
hh:MgC{l_2}(a\,mol),CuC{l_2}(b\,mol)\\
{P_1}:\\
2AgN{O_3} + MgC{l_2} \to 2AgCl + Mg{(N{O_3})_2}\\
2AgN{O_3} + CuC{l_2} \to 2AgCl + Cu{(N{O_3})_2}\\
{n_{AgCl}} = \dfrac{{28,7}}{{143,5}} = 0,2\,mol\\
\Rightarrow 0,5a \times 2 + 0,5b \times 2 = 0,2 \Leftrightarrow a + b = 0,2(1)\\
{P_2}:\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Mg{(OH)_2} \to MgO + {H_2}O\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,5a\,mol\\
{n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,5b\,mol\\
{n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,5a\,mol\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,5b\,mol\\
\Rightarrow 0,5a \times 40 + 0,5b \times 80 = 6 \Rightarrow 10a + 20b = 3(2)\\
(1),(2) \Rightarrow a = b = 0,1\\
{C_\% }MgC{l_2} = \dfrac{{0,1 \times 95}}{{100}} \times 100\% = 9,5\% \\
{C_\% }CuC{l_2} = \dfrac{{0,1 \times 135}}{{100}} \times 100\% = 13,5\%
\end{array}\)
