Em tham khảo nha:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
n{H_2} = \dfrac{V}{{22,4}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
56a + 27b = 11\\
a + 1,5b = 0,4\\
\Rightarrow a = 0,1;b = 0,2\\
\% mFe = \dfrac{{5,6}}{{11}} \times 100\% = 50,9\% \\
\% mAl = 100 - 50,9 = 49,1\% \\
c)\\
nHCl = 2n{H_2} = 0,8\,mol\\
VHCl = \dfrac{n}{{{C_M}}} = \dfrac{{0,8}}{2} = 0,4l\\
d)\\
{C_M}FeC{l_2} = \dfrac{{0,1}}{{0,4}} = 0,25M\\
{C_M}AlC{l_3} = \dfrac{{0,2}}{{0,4}} = 0,5M
\end{array}\)