Đáp án:

a) \({m_{Fe}} = 8,4{\text{ gam; }}{{\text{m}}_{Cu}} = 6,4{\text{ gam}}\)

b) \({{\text{V}}_{{H_2}}} = 7,28{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{}}2Fe + 3{H_2}O\)

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

\({m_{F{e_2}{O_3}}} = 20.60\%  = 12{\text{ gam; }}{{\text{m}}_{CuO}} = 20 - 12 = 8{\text{ gam}} \to {{\text{n}}_{F{e_2}{O_3}}} = \frac{{12}}{{56.2 + 16.3}} = 0,075{\text{ mol;}}{{\text{n}}_{CuO}} = \frac{8}{{64 + 16}} = 0,1{\text{ mol}}\)

\( \to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,075.2 = 0,15{\text{ mol; }}{{\text{n}}_{Cu}} = {n_{CuO}} = 0,1{\text{ mol}}\)

\( \to {m_{Fe}} = 0,15.56 = 8,4{\text{ gam; }}{{\text{m}}_{Cu}} = 0,1.64 = 6,4{\text{ gam}}\)

\({n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} + {n_{CuO}} = 0,075.3 + 0,1 = 0,325{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,325.22,4 = 7,28{\text{ lit}}\)