\(\begin{array}{l}
11)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Fe{(OH)_3} + 3HCl \to FeC{l_3} + 3{H_2}O\\
Fe{(OH)_3} + 3HN{O_3} \to Fe{(N{O_3})_3} + 3{H_2}O\\
12)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol \Rightarrow nFe = n{H_2} = 0,1\,mol\\
nS{O_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol \Rightarrow nCu = nS{O_2} = 0,15\,mol\\
\% mFe = \dfrac{{0,1 \times 56}}{{0,1 \times 56 + 0,15 \times 64}} \times 100\% = 36,84\% \\
\% mCu = 100 - 36,84 = 63,16\%
\end{array}\)
