Đáp án:

\({m_{HCl}} = 18,25{\text{ gam}}\)

\({m_{Cu}} = 9,6{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{{H_2}}} = 0,25.2 = 0,5{\text{ mol}}\)

\( \to {m_{HCl}} = 0,5.36,5 = 18,25{\text{ gam}}\)

Dùng \(H_2\) khử \(CuO\)

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

Ta có:

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

\( \to {n_{Cu}} = {n_{CuO}} = 0,15{\text{ mol}}\)

\( \to {m_{Cu}} = 0,15.64 = 9,6{\text{ gam}}\)