Em tham khảo nha :

\(\begin{array}{l}
25)A\\
F{e_3}{O_4} + 4CO \to 3Fe + 4C{O_2}\\
CuO + CO \to Cu + C{O_2}\\
{m_{F{e_3}{O_4}}} + {m_{CuO}} = 15,6g\\
{m_{F{e_3}{O_4}}} - {m_{CuO}} = 7,6g\\
 \Rightarrow {m_{F{e_3}{O_4}}} = 11,6g\\
{m_{CuO}} = 15,6 - 11,6 = 4g\\
{n_{F{e_3}{O_4}}} = \dfrac{m}{M} = \dfrac{{11,6}}{{232}} = 0,05mol\\
{n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{4}{{80}} = 0,05mol\\
{n_{Cu}} = {n_{CuO}} = 0,05mol\\
{m_{Cu}} = 0,05 \times 64 = 3,2g\\
26)A\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{8,4}}{{22,4}} = 0,375mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,375mol\\
{m_{Fe}} = 0,375 \times 56 = 21g\\
{n_{F{e_2}{O_3}}} = \dfrac{{{n_{Fe}}}}{2} = 0,1875mol\\
{m_{F{e_2}{O_3}}} = 0,1875 \times 160 = 30g\\
32)C\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}(1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(2)\\
{n_{{H_2}(1)}} = \dfrac{3}{2}{n_{Al}} = \dfrac{3}{2}a\,mol\\
{n_{{H_2}(2)}} = {n_{Fe}} = a\,mol\\
\dfrac{{{n_{{H_2}(1)}}}}{{{n_{{H_2}(2)}}}} = \dfrac{3}{2}:1 = \dfrac{3}{2}
\end{array}\)