Đáp án:

\( \% {m_{Al}}  = 39,35\% ;\% {m_{Zn}} = 60,65\% \)

\( {m_{dd}} = 101,48{\text{ gam}}\)

\( C{\% _{A{l_2}{{(S{O_4})}_3}}}  = 6,7\% \)

\(C{\% _{ZnS{O_4}}} = 6,35\% \)

Giải thích các bước giải:

 Gọi số mol \(Al;Zn\) lần lượt là \(x;y\)

\( \to 27x + 65y = 3,68{\text{ gam}}\)

Cho hỗn hợp tác dụng với \(H_2SO_4\)

\(2Al + 3{H_2}S{O_4}\xrightarrow{{{t^o}}}2A{l_2}{(S{O_4})_3} + 3{H_2}\)

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} + {n_{Zn}} = 1,5x + y = \frac{{2,24}}{{22,4}} = 0,1\)

Giải được: \(x=y=0,04\)

\( \to {m_{Al}} = 0,04.27 = 1,08{\text{ gam}}\)

\({m_{Zn}} = 0,04.65 = 2,6{\text{ gam}}\)

\( \to \% {m_{Al}} = \frac{{1,08}}{{3,68}} = 39,35\%   \)

\(\to \% {m_{Zn}} = 60,65\%\)

\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}}\)

\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{9,8}}{{10\% }} = 98{\text{ gam}}\)

BTKL:

\({m_{kl}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 3,68 + 98 = {m_{dd}} + 0,1.2\)

\( \to {m_{dd}} = 101,48{\text{ gam}}\)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,02{\text{ mol;}}{{\text{n}}_{ZnS{O_4}}} = {n_{Zn}} = 0,04{\text{ mol}}\)

\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,02.(27.2 + 96.3) = 6,84{\text{ gam}}\)

\({m_{ZnS{O_4}}} = 0,04.(65 + 96) = 6,44{\text{ gam}}\)

\( \to C{\% _{A{l_2}{{(S{O_4})}_3}}} = \frac{{6,84}}{{101,48}} = 6,7\% \)

\(C{\% _{ZnS{O_4}}} = \frac{{6,44}}{{101,48}} = 6,35\% \)