Đáp án:

 a) \(\% {m_{Al}} = 60\% ;\% {m_{Mg}} = 40\% \)

b) \({{\text{V}}_{S{O_2}}} = 14,112{\text{ lít}}\)

c) \({C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = 0,925M;{C_{M{\text{ }}{\text{Na HS}}{{\text{O}}_3}}} = 0,65M\)

Giải thích các bước giải:

Ta có: 

\({m_{Al}} + {m_{Mg}} = 12,6{\text{gam;}}{{\text{m}}_{Al}}:{m_{Mg}} = 3:2 \to {m_{Al}} = 7,56{\text{ gam; }}{{\text{m}}_{Mg}} = 5,04{\text{ gam}}\)

\(\% {m_{Al}} = \frac{{7,56}}{{12,6}} = 60\%  \to \% {m_{Mg}} = 40\% \)

Ta có: 

\({n_{Al}} = \frac{{7,56}}{{27}} = 0,28{\text{ mol;}}{{\text{n}}_{Mg}} = \frac{{5,04}}{{24}} = 0,21{\text{ mol}}\)

Bảo toàn e:

\(3{n_{Al}} + 2{n_{Mg}} = 2{n_{S{O_2}}} \to {n_{S{O_2}}} = \frac{{0,28.3 + 0,21.2}}{2} = 0,63{\text{ mol}} \to {{\text{V}}_{S{O_2}}} = 0,63.22,4 = 14,112{\text{ lít}}\)

Ta có: 

\({n_{NaOH}} = 0,4.2,5 = 1{\text{ mol}} \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{1}{{0,63}} = 1,59\) do vậy tạo ra hỗn hợp 2 muối.

Gọi số mol \(N{a_2}S{O_3};NaHS{O_3}\) lần lượt là x, y.

\( \to 2x + y = 1;x + y = 0,63 \to x = 0,37;y = 0,26\)

\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_3}}} = \frac{{0,37}}{{0,4}} = 0,925M;{C_{M{\text{ }}{\text{Na HS}}{{\text{O}}_3}}} = \frac{{0,26}}{{0,4}} = 0,65M\)