Đáp án:
\(\begin{array}{l}
a)\\
{m_{Cu(tt)}} = 11,52g\\
{m_{Fe(tt)}} = 10,08g\\
b){V_{{H_2}(dư)}} = 2,24l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{{H_2}}} = 0,6mol\\
{n_{CuO}} = 0,2mol\\
{n_{F{e_2}{O_3}}} = 0,1mol\\
\to {n_{{H_2}(1)}} = {n_{CuO}} = 0,2mol\\
\to {n_{{H_2}(2)}} = 3{n_{F{e_2}{O_3}}} = 0,3mol\\
\to {n_{{H_2}(du)}} = 0,1mol\\
a)\\
{n_{Cu}} = {n_{CuO}} = 0,2mol \to {m_{Cu}} = 12,8g\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2mol \to {m_{Fe}} = 11,2g\\
\to {m_{Cu(tt)}} = \dfrac{{12,8 \times 90}}{{100}} = 11,52g\\
\to {m_{Fe(tt)}} = \dfrac{{11,2 \times 90}}{{100}} = 10,08g\\
b)\\
{V_{{H_2}(dư)}} = 2,24l
\end{array}\)
