1) giả sử hh chỉ gồm Al
\(\begin{array}{l}
nAl = \dfrac{{22}}{{27}} = 0,81\\
nHCl = 8,5 \times 0,3 = 2,55\,mol\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
\dfrac{{0,81}}{2} < \dfrac{{2,55}}{6}
\end{array}\)
=> Al tan hết
Giả sử hh chỉ gồm Fe
\(\begin{array}{l}
nFe = \dfrac{{22}}{{56}} = 0,39\\
nHCl = 8,5 \times 0,3 = 2,55\,mol\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
\dfrac{{0,39}}{1} < \dfrac{{2,55}}{2}
\end{array}\)
=> Fe tan hết
=> X tan hết
2)
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
n{H_2} = \dfrac{{17,92}}{{22,4}} = 0,8\,mol\\
hhX:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 22\\
1,5a + b = 0,8\\
= > a = 0,4;b = 0,2\\
nHClspu = 2,55 - 1,6 = 0,95\,mol\\
CMHCl = \dfrac{{0,95}}{{8,5}} = 0,11M\\
CMFeC{l_2} = \dfrac{{0,2}}{{8,5}} = 0,024M\\
CMAlC{l_3} = \dfrac{{0,4}}{{8,5}} = 0,047M
\end{array}\)
