Đáp án:

 \(\begin{array}{l}
b)\\
{V_{NaOH}} = 1,8l = 1800ml\\
c)\\
{C_{{M_{NaCl}}}} = 0,214M\\
d)\\
{m_{NaCl}} = 26,325g\\
{V_{{H_2}S{O_4}}} = 96,7ml
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
2Fe{(OH)_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 6{H_2}O\\
b)\\
{n_{FeC{l_3}}} = {C_M} \times V = 0,3 \times 0,5 = 0,15mol\\
{n_{NaOH}} = 3{n_{FeC{l_3}}} = 0,45mol\\
{V_{NaOH}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,45}}{{0,25}} = 1,8l = 1800ml\\
c)\\
{n_{NaCl}} = 3{n_{FeC{l_3}}} = 0,45mol\\
{C_{{M_{NaCl}}}} = \dfrac{n}{V} = \dfrac{{0,45}}{{0,3 + 1,8}} = 0,214M\\
d)\\
{m_{NaCl}} = n \times M = 0,45 \times 58,5 = 26,325g\\
{n_{Fe{{(OH)}_3}}} = {n_{FeC{l_3}}} = 0,15mol\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Fe({{(OH)}_3}}} = 0,225mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,225 \times 98 = 22,05g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{22,05 \times 100}}{{20}} = 110,25g\\
{V_{{H_2}S{O_4}}} = \dfrac{m}{d} = \dfrac{{110,25}}{{1,14}} = 96,7ml
\end{array}\)