a) mNaOH=400.10/100=40g

nNaOH=40/40=1mol

mFe2(SO4)3=200.8/100=16g

nFe2(SO4)3=16/400=0,04mol

Fe2(SO4)3+6NaOH→2Fe(OH)3+3Na2SO4

0,04              0,24         0,08           0,12

mFe(OH)3=0,08.107=8,56g

b) mdd sau pứ=400+200-8,56=591,44g

mNa2SO4=0,12.142=17,04g

C%Na2SO4=17,04/591,44.100%=2,88%

nNaOH dư=1-0,24=0,76mol

mNaOH dư=0,76.40=30,4g

C%NaOH dư=30,4/591,44.100%=5,14%

Em tham khảo nha :

\(\begin{array}{l}
a)6NaOH + F{e_2}{(S{O_4})_3} \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
{m_{NaOH}} = \dfrac{{400 \times 10}}{{100}} = 40g\\
{n_{NaOH}} = \dfrac{{40}}{{40}} = 1mol\\
{m_{F{e_2}S{O_4}}} = \dfrac{{200 \times 8}}{{100}} = 16g\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{16}}{{400}} = 0,04mol\\
\dfrac{1}{6} > \dfrac{{0,04}}{1} \Rightarrow NaOH\text{ dư}\\
{n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{{(S{O_4})}_2}}} = 0,08mol\\
{m_{Fe{{(OH)}_3}}} = 0,08 \times 107 = 8,56g\\
b)\\
{n_{NaO{H_d}}} = 1 - 0,04 \times 6 = 0,76mol\\
{m_{NaO{H_d}}} = 0,76 \times 40 = 30,4g\\
{n_{N{a_2}S{O_4}}} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,12mol\\
{m_{N{a_2}S{O_4}}} = 0,12 \times 142 = 17,04g\\
C{\% _{NaO{H_d}}} = \dfrac{{30,4}}{{400 + 200 - 8,56}} \times 100\%  = 5,14\% \\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{17,04}}{{400 + 200 - 8,56}} \times 100\%  = 2,88\% 
\end{array}\)