Giúp mình với hôm nay mình thì rồi

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
\frac{3}{2}a + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
\% Al = \dfrac{{5,4}}{{11}} \times 100\%  = 49,1\% \\
\% Fe = 100 - 49,1 = 50,9\% \\
c)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Fe}} = 0,8mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,8}}{{0,4}} = 2M\\
d)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,2mol\\
{C_{{M_{AlC{l_3}}}}} = \dfrac{{0,2}}{{0,4}} = 0,5M\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,1}}{{0,4}} = 0,25M
\end{array}\)