Giải thích các bước giải:

Em tham khảo nha :

\(\begin{array}{l}
a)\\
hh:F{e_2}{O_3}(a\,mol),FeO(b\,mol)\\
\left\{ \begin{array}{l}
a - b = 0\\
160a + 72b = 11,6
\end{array} \right.\\
 \Rightarrow a = 0,05;b = 0,05\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{HCl}} = 0,3 \times 2 = 0,6mol\\
{n_{HC{l_{pu}}}} = 6{n_{F{e_2}{O_3}}} + 2{n_{FeO}} = 0,4mol\\
{n_{HC{l_d}}} = 0,6 - 0,4 = 0,2mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
{n_{FeC{l_2}}} = {n_{FeO}} = 0,05mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,2}}{{0,3}} = 0,667M\\
{C_{{M_{FeC{l_3}}}}} = \dfrac{{0,1}}{{0,3}} = 0,333M\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,05}}{{0,3}} = 0,1667M\\
b)\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = 3{n_{FeC{l_3}}} + 2{n_{FeC{l_2}}} + {n_{HCl}} = 0,6mol\\
{V_{NaOH}} = \dfrac{{0,6}}{{1,5}} = 0,4l = 400ml
\end{array}\)