Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
b)\\
nMg = \dfrac{m}{M} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
nAl = \dfrac{m}{M} = \dfrac{{8,1}}{{27}} = 0,3\,mol\\
n{H_2} = 0,1 + 0,45 = 0,55\,mol\\
V{H_2} = 0,55 \times 22,4 = 12,32l\\
nHCl = 2n{H_2} = 0,55 \times 2 = 1,1\,mol\\
m{\rm{dd}}HCl = \dfrac{{1,1 \times 36,5}}{{7,3\% }} = 550g\\
2)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
nA{l_2}{O_3} = \dfrac{m}{M} = \dfrac{{15,3}}{{102}} = 0,15\,mol\\
n{H_2}S{O_4} = \dfrac{{250 \times 9,8\% }}{{98}} = 0,25\,mol
\end{array}\)

Em xem lại đề câu 2 nha