7)

Phản ứng xảy ra:

\(2{\text{A}}l + 6HCl\xrightarrow{{}}2{\text{A}}lC{l_3} + 3{H_2}\)

Ta có:

\({n_{Al}} = \frac{{81}}{{27}} = 3{\text{ mol = }}{{\text{n}}_{AlC{l_3}}}\)

\( \to {m_{AlC{l_3}}} = 3.(27 + 35,5.3) = 400,5{\text{ gam}}\)

\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} = \frac{3}{2}.3 = 4,5{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 4,5.22,4 = 10,08{\text{ lít}}\)

Dẫn lượng \(H_2\) trên qua \(CuO\)

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

\( \to {n_{CuO}} = \frac{{16}}{{64 + 16}} = 0,2{\text{ mol < }}{{\text{n}}_{{H_2}}}\)

Vậy \(H_2\) dư

\( \to {n_{Cu}} = {n_{CuO}} = 0,2{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,2.64 = 12,8{\text{ gam}}\)

8)

Phản ứng xảy ra:

\(2Na + 2{H_2}O\xrightarrow{{{t^o}}}2NaOH + {H_2}\)

Ra có:

\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {{\text{n}}_{NaOH}} = 2{n_{{H_2}}} = 0,3.2 = 0,6{\text{ mol}}\)

\( \to {m_{NaOH}} = 0,6.40 = 24{\text{ gam}}\)

Câu \(b\) cho \(H_2\) rồi bắt tính làm gì nữa

Điều chế lượng \(H_2\) trên

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

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

\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam}}\)