Em tham khảo nha :

\(\begin{array}{l}
a)\\
\text{Giả sử hôn hợp chỉ có Al}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{6,3}}{{27}} = 0,2333mol\\
{m_{HCl}} = \dfrac{{200 \times 14,6}}{{100}} = 29,2g\\
{n_{HCl}} = \dfrac{{29,2}}{{36,5}} = 0,8mol\\
\dfrac{{0,2333}}{2} < \dfrac{{0,8}}{6} \Rightarrow HCl\text{ dư}\\
b)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 6,3\\
\dfrac{3}{2}a + b = 0,3
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,15\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
\% Al = \dfrac{{2,7}}{{6,3}} \times 100\%  = 42,9\% \\
\% Mg = 100 - 42,9 = 57,1\% \\
c)\\
{V_{HCl}} = \dfrac{{200}}{{1,1}} = 181,818ml = 0,1818l\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,15mol\\
{C_{{M_{MgC{l_2}}}}} = \dfrac{{0,15}}{{0,1818}} = 0,825M\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,1}}{{0,1818}} = 0,55M\\
{n_{HC{l_d}}} = 0,8 - 0,1 \times 3 - 0,15 \times 2 = 0,2mol\\
{C_{{M_{HC{l_d}}}}} = \dfrac{{0,2}}{{0,1818}} = 1,1M
\end{array}\)