Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{4,368}}{{22,4}} = 0,195mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
24a + 27b = 3,87(1)\\
a + \dfrac{3}{2}b = 0,195(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,06mol;b = 0,09mol\\
{m_{Mg}} = 0,06 \times 24 = 1,44g\\
\% Mg = \dfrac{{1,44}}{{3,87}} \times 100\% = 37,2\% \\
\% Al = 100 - 37,2 = 62,8\% \\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,06mol\\
{C_{{M_{MgC{l_2}}}}} = \dfrac{{0,06}}{{0,5}} = 0,12M\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,09mol\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,09}}{{0,5}} = 0,18M\\
{n_{HC{l_{pu}}}} = 2{n_{Mg}} + 3{n_{Al}} = 0,39mol\\
{n_{HCl}} = 0,5 \times 1 = 0,5mol\\
{n_{HC{l_d}}} = 0,5 - 0,39 = 0,11mol\\
{C_{{M_{HC{l_d}}}}} = \dfrac{{0,11}}{{0,5}} = 0,22M
\end{array}\)