Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
56a + 24b = 10,4\\
a + b = 0,3
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{10,4}} \times 100\% = 53,85\% \\
\% Cu = 100 - 53,85 = 46,15\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{21,9 \times 100}}{{10}} = 219g\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2mol\\
{m_{MgC{l_2}}} = 0,2 \times 95 = 19g\\
C{\% _{FeC{l_2}}} = \dfrac{{12,7}}{{10,4 + 219 - 0,3 \times 2}} \times 100\% = 5,55\% \\
C{\% _{MgC{l_2}}} = \dfrac{{19}}{{10,4 + 219 - 0,3 \times 2}} \times 100\% = 8,3\%
\end{array}\)
