Đáp án:
\(\begin{array}{l}
a)\\
x = 1\\
b)\\
{C_\% }Ba{(OH)_2} \text{ dư }= 13,45\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,3 \times 2 = 0,6\,mol\\
{n_{BaS{O_4}}} = \dfrac{{46,6}}{{233}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,2\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,2}}{{0,2}} = 1M\\
b)\\
{m_{{\rm{dd}}Ba{{(OH)}_2}}} = 300 \times 1,05 = 315g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 1,2 \times 200 = 240g\\
{m_{{\rm{dd}}B}} = 315 + 240 - 46,6 = 508,4g\\
{n_{Ba{{(OH)}_2}}} \text{ dư }= 0,6 - 0,2 = 0,4\,mol\\
{C_\% }Ba{(OH)_2} \text{ dư }= \dfrac{{0,4 \times 171}}{{508,4}} \times 100\% = 13,45\%
\end{array}\)