Em tham khảo nha:

\(\begin{array}{l}
5)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
nFe = nFeC{l_2} = n{H_2} = 0,1\,mol\\
mAlC{l_3} = 39,4 - 0,1 \times 127 = 26,7g\\
nAlC{l_3} = \dfrac{{26,7}}{{133,5}} = 0,2\,mol \Rightarrow nA{l_2}{O_3} = 0,1\,mol\\
mFe = 0,1 \times 56 = 5,6g\\
mA{l_2}{O_3} = 0,1 \times 102 = 10,2g\\
6)\\
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
n{H_2} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
nAl = \dfrac{{0,45 \times 2}}{3} = 0,3\,mol\\
mAl = 0,3 \times 27 = 8,1(g)\\
mAg = 18,9 - 8,1 = 10,8g\\
b)\\
nHCl = 2n{H_2} = 0,9\,mol\\
VHCl = \dfrac{{0,9}}{{0,4}} = 2,25l\\
c)\\
nAlC{l_3} = nAl = 0,3\,mol\\
{C_M}AlC{l_3} = \dfrac{{0,3}}{{2,25}} = 0,13M
\end{array}\)