\(\begin{array}{l}
2)\\
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
n{H_2} = \dfrac{{15,68}}{{22,4}} = 0,7\,mol\\
hh:Al(a\,mol),Zn(b\,mol)\\
27a + 65b = 31,4\\
1,5a + b = 0,7\\
\Rightarrow a = 0,2;b = 0,4\\
\% mAl = \dfrac{{0,2 \times 27}}{{31,4}} \times 100\% = 17,2\% \\
\% mZn = 100 - 17,2 = 82,8\% \\
b)\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
nS{O_2} = nZn = 0,4\,mol\\
\Rightarrow VS{O_2} = 0,4 \times 22,4 = 8,96l\\
3)\\
nS{O_2} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
nNaOH = 0,5 \times 1 = 0,5\,mol\\
T = \dfrac{{nNaOH}}{{nS{O_2}}} = \dfrac{{0,5}}{{0,4}} = 1,25 \Rightarrow \text{ Tạo cả 2 muối}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O(1)\\
N{a_2}S{O_3} + S{O_2} + {H_2}O \to NaHS{O_3}(2)\\
nN{a_2}S{O_3}(1) = nS{O_2}(1) = 0,25\,mol\\
nN{a_2}S{O_3}(2) = nS{O_2}(2) = 0,4 - 0,25 = 0,15\,mol\\
nN{a_2}S{O_3} = 0,25 - 0,15 = 0,1\,mol\\
nNaHS{O_3} = 0,15 \times 2 = 0,3\,mol\\
{C_M}N{a_2}S{O_3} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
{C_M}NaHS{O_3} = \dfrac{{0,3}}{{0,5}} = 0,6M
\end{array}\)
