Đáp án:

a) \({\text{\% }}{{\text{m}}_{Al}} = 69,68\% ;\% {m_{Ag}} = 30,32\% \)

b) \({m_{A{l_2}{{(S{O_4})}_3}}} = 89,4{\text{ gam; }}{{\text{m}}_{A{g_2}S{O_4}}} = 8,86{\text{ gam}}\)

c)  \({{\text{V}}_{{H_2}S{O_4}}} = 0,271{\text{ lít}}\)

Giải thích các bước giải:

 Gọi số mol Al là x; Ag là y.

\( \to 27x + 108y = 20,25\)

Cho hỗn hợp tác dụng với axit sunfuric đặc nóng.

\(2Al + 6{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)

\(2Ag + 2{H_2}S{O_4}\xrightarrow{{}}A{g_2}S{O_4} + S{O_2} + 2{H_2}O\)

Ta có: \({n_{S{O_2}}} = \frac{3}{2}{n_{Al}} + \frac{1}{2}{n_{Ag}} = 1,5x + 0,5y = \frac{{18,2}}{{22,4}} = 0,8125{\text{ mol}}\)

Giải được: \(x = \frac{{23}}{{44}};y = \frac{5}{{88}}\)

\( \to {m_{Al}} = \frac{{23}}{{44}}.27 = 14,11{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Al}} = \frac{{14,11}}{{20,25}} = 69,68\%  \to \% {m_{Ag}} = 30,32\% \)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = \frac{{23}}{{88}}{\text{ mol; }}{{\text{n}}_{A{g_2}S{O_4}}} = \frac{1}{2}{n_{Ag}} = \frac{5}{{176}}{\text{ mol}}\)

\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = \frac{{23}}{{88}}.(27.2 + 96.3) = 89,4{\text{ gam; }}{{\text{m}}_{A{g_2}S{O_4}}} = \frac{5}{{176}}.(108.2 + 96) = 8,86{\text{ gam}}\)

\({n_{{H_2}S{O_4}}} = 2{n_{S{O_2}}} = 0,8125.2 = 1,625{\text{ mol}} \to {{\text{V}}_{{H_2}S{O_4}}} = \frac{{1,625}}{6} = 0,271{\text{ lít}}\)