Giải thích các bước giải:

\(\begin{array}{l}
6NaOH + F{e_2}{(S{O_4})_3} \to 3N{a_2}S{O_4} + 2Fe{(OH)_3}\\
{m_{NaOH}} = \dfrac{{200 \times 8\% }}{{100\% }} = 16g \to {n_{NaOH}} = 0,4mol\\
a)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{6}{n_{NaOH}} = 0,067mol \to {m_{F{e_2}{{(S{O_4})}_3}}} = 26,8g\\
 \to C{\% _{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{26,8}}{{300}} \times 100\%  = 8,93\% \\
b)\\
{n_{Fe{{(OH)}_3}}} = \dfrac{1}{3}{n_{NaOH}} = 0,13mol \to {m_{Fe{{(OH)}_3}}} = 13,91g\\
c)\\
{n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol \to {m_{N{a_2}S{O_4}}} = 28,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}NaOH}} + {m_{{\rm{dd}}F{e_2}{{(S{O_4})}_3}}} - {m_{Fe{{(OH)}_3}}} = 486,09g\\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{28,4}}{{486,09}} \times 100\%  = 5,84\% \\
d)\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_3}}} = 0,065mol \to {m_{F{e_2}{O_3}}} = 10,4g
\end{array}\)