1)

Phản ứng xảy ra:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có:

\({n_{Fe}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{HCl}} = 2{n_{{H_2}}} = 0,1.2 = 0,2{\text{ mol}}\)

\( \to {m_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{7,3}}{{100}} = 7,3\% \)

Khử \(CuO\)

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

\({n_{CuO}} = \frac{8}{{64 + 16}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}}}\)

Vậy phản ứng vừa đủ

\( \to {n_{Cu}} = {n_{CuO}} = 0,1{\text{ mol}}\)

\( \to {m_{Cu}} = 0,1.64 = 6,4{\text{ gam}}\)

2)

Phản ứng xảy ra:

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

\({n_{Al}} = \frac{{8,1}}{{27}} = 0,3{\text{ mol}}\)

\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,45{\text{ mol}}\)

\( \to V = {V_{{H_2}}} = 0,45.22,4 = 10,08{\text{ lít}}\)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,15{\text{ mol}}\)

\( \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,15.(27.2 + 96.3) = 51,3{\text{ gam}}\)