1)
Phản ứng xảy ra:
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,3{\text{ mol}} \to {\text{V = }}{{\text{V}}_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
2)
Gọi số mol Zn và Fe lần lượt là x, y.
\( \to 65x + 56y = 18,6\)
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\( \to {n_{{H_2}}} = {n_{Zn}} + {n_{Fe}} = x + y = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}}\)
Giải được x=0,2; y=0,1.
\( \to {m_{Zn}} = 0,2.65 = 13{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Zn}} = \frac{{13}}{{18,6}} = 69,9\% \to \% {m_{Fe}} = 30,1\% \)
\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{C}}_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,3}}{{0,1}} = 3M\)