Đáp án:

\(\begin{array}{l}
a)\\
{m_{Fe{{(OH)}_3}}} = 21,4g\\
b)\\
C{\% _{N{a_2}S{O_4}}} = 15,29\% \\
C{\% _{F{e_2}{{(S{O_4})}_3}}} = 7,18\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
6NaOH + F{e_2}{(S{O_4})_3} \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
{m_{NaOH}} = \dfrac{{150 \times 16}}{{100}} = 24g\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{24}}{{40}} = 0,6mol\\
{m_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{150 \times 40}}{{100}} = 60g\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{m}{M} = \dfrac{{60}}{{400}} = 0,15mol\\
\dfrac{{0,6}}{6} < \dfrac{{0,15}}{1} \Rightarrow F{e_2}{(S{O_4})_3}\text{ dư}\\
\\
{n_{Fe{{(OH)}_3}}} = \dfrac{2}{6}{n_{NaOH}} = 0,2mol\\
{m_{Fe{{(OH)}_3}}} = n \times M = 0,2 \times 107 = 21,4g\\
b)\\
{m_{ddspu}} = 150 + 150 - 21,4 = 278,6g\\
{n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,3mol\\
{m_{N{a_2}S{O_4}}} = n \times M = 0,3 \times 142 = 42,6g\\
{n_{F{e_2}{{(S{O_4})}_3}d}} = 0,15 - \dfrac{{0,6}}{6} = 0,05mol\\
{m_{F{e_2}{{(S{O_4})}_3}d}} = n \times M = 0,05 \times 400 = 20g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{42,6}}{{278,6}} \times 100\%  = 15,29\% \\
C{\% _{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{20}}{{278,6}} \times 100\%  = 7,18\% 
\end{array}\)