Đáp án:

 a) A là Al

b) \({{\text{V}}_{HCl}} = 0,09{\text{ lít}}\)

c) \({{\text{m}}_{Cu}} = 5,76{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2A + 6HCl\xrightarrow{{}}2AC{l_3} + 3{H_2}\)

Ta có: \({n_A} = {n_{AlC{l_3}}} \to \frac{{1,62}}{A} = \frac{{8,01}}{{A + 35,5.3}} \to A = 27\)

Vậy A là Al.

Ta có: 

\({n_A} = \frac{{1,62}}{{27}} = 0,06{\text{ mol}} \to {{\text{n}}_{HCl}} = 3{n_{Al}} = 0,18{\text{ mol}} \to {{\text{V}}_{HCl}} = \frac{{0,18}}{2} = 0,09{\text{ lit}}\)

\({n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,09{\text{ mol}}\)

\(CuO + {H_2}\xrightarrow{{}}Cu + {H_2}O\)

\({n_{CuO}} = \frac{{12}}{{64 + 16}} = 0,15{\text{ mol > }}{{\text{n}}_{{H_2}}}\)

Do vậy CuO dư.

\(\to {n_{Cu}} = {n_{{H_2}}} = 0,09{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,09.64 = 5,76{\text{ gam}}\)