Em tham khảo nha :
\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{m_{NaOH}} = \dfrac{{200 \times 16}}{{100}} = 32g\\
{n_{NaOH}} = \dfrac{{32}}{{40}} = 0,8mol\\
{m_{Ba{{(OH)}_2}}} = \dfrac{{200 \times 17,1}}{{100}} = 34,2g\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{34,2}}{{171}} = 0,2mol\\
{n_{{H_2}S{O_{{4_{pu}}}}}} = \dfrac{{{n_{NaOH}}}}{2} + {n_{Ba{{(OH)}_2}}} = 0,6mol\\
{n_{{H_2}S{O_4}d}} = \dfrac{{0,6 \times 20}}{{100}} = 0,12mol\\
{n_{{H_2}S{O_4}}} = 0,6 + 0,12 = 0,72mol\\
{m_{{H_2}S{O_4}}} = 0,72 \times 98 = 70,56g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{70,56 \times 100}}{{9,8}} = 720g\\
{n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,2mol\\
{m_{BaS{O_4}}} = 0,2 \times 233 = 46,6g\\
b)\\
{m_{{\rm{dd}}spu}} = 200 + 720 - 46,6 = 873,4g\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,4mol\\
{m_{N{a_2}S{O_4}}} = 0,4 \times 142 = 56,8g\\
{m_{{H_2}S{O_4}d}} = 0,12 \times 98 = 11,76g\\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{56,8}}{{873,4}} \times 100\% = 6,5\% \\
C{\% _{{H_2}S{O_4}}} = \dfrac{{11,76}}{{873,4}} \times 100\% = 1,35\%
\end{array}\)
