Em tham khảo nha:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Fe{(OH)_3} + 3HCl \to FeC{l_3} + 3{H_2}O\\
{n_{HCl}} = 0,35 \times 1 = 0,35\,mol\\
hh:Fe(a\,mol),F{e_2}{O_3}(b\,mol)\\
56a + 107b = 10,95\\
2a + 3b = 0,35\\
 \Rightarrow a = 0,1;b = 0,05\\
\% {n_{Fe}} = \dfrac{{0,1}}{{0,1 + 0,05}} \times 100\%  = 66,67\% \\
\% {n_{F{e_2}{O_3}}} = 100 - 66,67 = 33,33\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol \Rightarrow {C_M}FeC{l_2} = \dfrac{{0,1}}{{0,35}} \approx 0,286M\\
{n_{FeC{l_3}}} = {n_{Fe{{(OH)}_3}}} = 0,05\,mol \Rightarrow {C_M}FeC{l_3} = \dfrac{{0,05}}{{0,35}} \approx 0,143M\\
c)\\
FeC{l_2} + 2KOH \to Fe{(OH)_2} + 2KCl\\
FeC{l_3} + 3KOH \to Fe{(OH)_3} + 3KCl\\
{n_{KOH}} = 0,1 \times 2 + 0,05 \times 3 = 0,35\,mol\\
{V_{KOH}} = \dfrac{{0,35}}{{1,2}} = 0,29167(l)
\end{array}\)