Đáp án:

 b) 50% , 50%

c) 5600 l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Cu}} = 12 - 0,1 \times 56 = 6,4g\\
{n_{Cu}} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{n_{CuO}} = {n_{Cu}} = 0,1\,mol\\
\% {m_{F{e_2}{O_3}}} = \dfrac{{0,05 \times 160}}{{0,05 \times 160 + 0,1 \times 80}} \times 100\%  = 50\% \\
\% {m_{CuO}} = 100 - 50 = 50\% \\
c)\\
{n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} + {n_{CuO}} = 3 \times 0,05 + 0,1 = 0,25\,mol\\
{V_{{H_2}}} = 0,25 \times 22,4 = 5,6l = 5600ml
\end{array}\)