Em tham khảo nha :

\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,3 \times 1 = 0,3mol\\
{n_{{H_2}S{O_4}}} = 0,2 \times 0,8 = 0,16mol\\
\dfrac{{0,3}}{2} < \dfrac{{0,16}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}d}} = 0,16 - \dfrac{{0,3}}{2} = 0,01mol\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,15mol\\
{C_{{M_{{H_2}S{O_4}d}}}} = \dfrac{{0,01}}{{0,3 + 0,2}} = 0,02M\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{{0,15}}{{0,3 + 0,2}} = 0,3M\\
b)\\
Ba{(OH)_2} + N{a_2}S{O_4} \to 2NaOH + BaS{O_4}\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = {n_{N{a_2}S{O_4}}} + {n_{{H_2}S{O_4}d}} = 0,15 + 0,01 = 0,16mol\\
{m_{Ba{{(OH)}_2}}} = 0,16 \times 171 = 27,36g\\
{m_{{\rm{dd}}Ba{{(OH)}_2}}} = \dfrac{{27,36 \times 100}}{{10}} = 273,6g\\
{n_{BaS{O_4}}} = {n_{N{a_2}S{O_4}}} + {n_{{H_2}S{O_4}d}} = 0,16mol\\
{m_{BaS{O_4}}} = 0,16 \times 233 = 37,28g
\end{array}\)

                  2NaOH + H2SO4 -> Na2SO4 + H2O

ban đầu      0.3             0.16

phản ứng:    0.3            0.15           0.15

sau p/ư        0               0.01           0.15

V dd sau pư = 0.3 + 0.2 = 0.5 l

=> CM Na2SO4 = 0.15/0.5 = 0.3 M

Ba(OH)2 + H2SO4 -> BaSO4 + 2H2O

0.01            0.01           0.01

Ba(OH)2 + Na2SO4 -> BaSO4 + 2NaOH

0.15              0.15          0.15

nBa(OH)2 = 0.16 mol = nBaSO4

-> m ktua BaSO4 = 0.16 x 233 = 37.28g

m dd Ba(OH)2 = 0.16 x 171 x 100/10 = 273.6g