Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(1)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O(2)\\
{m_{HCl}} = \dfrac{{182,5 \times 20\% }}{{100\% }} = 36,5g\\
\to {n_{HCl}} = 1mol\\
\to {n_{HCl(1)}} = 2{n_{Fe}} = 0,3mol\\
\to {n_{HCl(2)}} = 6{n_{Fe}} = 0,6mol\\
\to {n_{HCl(dư)}} = 0,1mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,2mol\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl(3)\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl(4)\\
NaOH + HCl \to NaCl + {H_2}O(5)\\
b)\\
{m_{NaOH}} = \dfrac{{192 \times 25\% }}{{100\% }} = 48g\\
\to {n_{NaOH}} = 1,2g\\
\to {n_{NaOH(3)}} = 2{n_{FeC{l_3}}} = 0,3mol\\
\to {n_{NaOH(4)}} = 3{n_{FeC{l_3}}} = 0,6mol\\
\to {n_{NaOH(5)}} = {n_{HCl}} = 0,1mol\\
\to {n_{NaOH(dư)}} = 0,2mol\\
\to {m_{NaOH(dư)}} = 0,2 \times 40 = 8g\\
\to {n_{NaCl}} = 2{n_{FeC{l_2}}} + {3_{FeC{l_3}}} + {n_{HCl}} = 1mol\\
\to {m_{NaCl}} = 58,5 = 58,5g\\
{n_{Fe{{(OH)}_2}}} = {n_{Fe}} = 0,15mol\\
{n_{Fe{{(OH)}_3}}} = {n_{F{e_2}{O_3}}} = 0,1mol\\
\to {m_{ddB}} = {m_{ddA}} + {m_{ddNaOH}} - ({m_{Fe{{(OH)}_2}}} + {m_{Fe{{(OH)}_3}}})\\
\to {m_{ddB}} = ({m_{hỗnhợp(Fe,F{e_2}{O_3})}} + {m_{ddHCl}} - {m_{{H_2}}}) + {m_{ddNaOH}} - ({m_{Fe{{(OH)}_2}}} + {m_{Fe{{(OH)}_3}}})\\
\to {m_{ddB}} = (24,4 + 182,5 - 0,3) + 192 - (13,5 + 10,7) = 374,4g\\
\to C{\% _{NaCl}} = \dfrac{{58,5}}{{374,4}} \times 100\% = 15,625\% \\
\to C{\% _{NaOH(dư)}} = \dfrac{8}{{374,4}} \times 100\% = 2,14\%
\end{array}\)
