Đáp án:

\(\begin{array}{l}
a)\\
{m_{Fe}} = 8,4g\\
{m_{Cu}} = 6,4g\\
b)\\
{V_{{H_2}}} = 7,28l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{m_{F{e_2}{O_3}}} = 20 \times 60\%  = 12g\\
{n_{F{e_2}{O_3}}} = \dfrac{{12}}{{160}} = 0,075\,mol\\
{m_{CuO}} = 20 - 12 = 8g\\
{n_{CuO}} = \dfrac{8}{{80}} = 0,1\,mol\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 2 \times 0,075 = 0,15\,mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{n_{Cu}} = {n_{CuO}} = 0,1\,mol\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
b)\\
{n_{{H_2}}} = 0,075 \times 3 + 0,1 = 0,325\,mol\\
{V_{{H_2}}} = 0,325 \times 22,4 = 7,28l
\end{array}\)