\(\begin{array}{l}
1)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
nAl = \frac{{1,08}}{{27}} = 0,04\,mol\\
= > n{H_2} = 0,06\,mol\\
V{H_2} = 0,06 \times 22,4 = 1,44l\\
b)\\
MO + {H_2} \to M + {H_2}O\\
MMO = \frac{{4,8}}{{0,06}} = 80g/mol\\
= > MM = 64g/mol\\
= > M:Cu = > CTHH:CuO\\
2)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
a)\\
\% mZn = \frac{{6,5}}{{11,9}} \times 100\% = 54,62\% \\
\% mAl = 100 - 54,62 = 45,38\% \\
b)\\
nZn = 0,1\,mol\\
nAl = 0,2\,mol\\
= > n{H_2} = 0,4\,mol\\
V{H_2} = 0,4 \times 22,4 = 8,96l
\end{array}\)