Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
24a + 56b = 10,4
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{Mg}} = 0,2 \times 24 = 4,8g\\
\% Mg = \dfrac{{4,8}}{{10,4}} \times 100\% = 46,15\% \\
\% Fe = 100 - 46,15 = 53,85\% \\
b)\\
{m_{HCl}} = \dfrac{{400 \times 7,3}}{{100}} = 29,2g\\
{n_{HCl}} = \dfrac{{29,2}}{{36,5}} = 0,8mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2mol\\
{m_{MgC{l_2}}} = 0,2 \times 95 = 19g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{Fe}} - 2{n_{Mg}} = 0,2mol\\
{m_{HC{l_d}}} = 0,2 \times 36,5 = 7,3g\\
{m_{{\rm{dd}}spu}} = 10,4 + 400 - 0,3 \times 2 = 409,8g\\
C{\% _{HC{l_d}}} = \dfrac{{7,3}}{{409,8}} \times 100\% = 1,78\% \\
C{\% _{MgC{l_2}}} = \dfrac{{19}}{{409,8}} \times 100\% = 4,64\% \\
C{\% _{FeC{l_2}}} = \dfrac{{12,7}}{{409,8}} \times 100\% = 3,1\%
\end{array}\)
