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:
\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
{m_{F{e_2}{O_3}}} = \dfrac{{20 \times 60}}{{100}} = 12g\\
{m_{CuO}} = 20 - 12 = 8g\\
{n_{F{e_2}{O_3}}} = \dfrac{{12}}{{160}} = 0,075mol\\
{n_{CuO}} = \dfrac{8}{{80}} = 0,1mol\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{n_{Cu}} = {n_{CuO}} = 0,1mol\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
b)\\
{n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} + {n_{CuO}} = 0,325mol\\
{V_{{H_2}}} = 0,325 \times 22,4 = 7,28l
\end{array}\)