Em tham khảo nha :
\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
{m_{F{e_2}{O_3}}} = 2{m_{CuO}}\\
hh:F{e_2}{O_3}(a\,mol);CuO(b\,mol)\\
\left\{ \begin{array}{l}
160a - 160b = 0\\
112a + 64b = 17,6
\end{array} \right.\\
\Rightarrow a = ,1;b = 0,1\\
{m_{F{e_2}{O_3}}} = 0,1 \times 160 = 16g\\
{m_{CuO}} = 0,1 \times 80 = 8g\\
b)\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2mol\\
{m_{Fe}} = 0,2 \times 56 = 11,2g\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{m_{FeC{l_2}}} = 0,2 \times 127 = 25,4g\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
C{\% _{FeC{l_2}}} = \dfrac{{25,4}}{{11,2 + 100 - 0,2 \times 2}} \times 100\% = 22,92\%
\end{array}\)