Em tham khảo nha :
\(\begin{array}{l}
a)\\
S{O_2} + Ba{(OH)_2} \to BaS{O_3} + {H_2}O\\
b)\\
{n_{S{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{BaS{O_3}}} = {n_{S{O_2}}} = 0,15mol\\
{m_{BaS{O_3}}} = 0,15 \times 217 = 32,55g\\
{n_{Ba{{(OH)}_2}}} = 0,4 \times 1 = 0,4mol\\
{n_{Ba{{(OH)}_2}d}} = 0,4 - 0,15 = 0,25mol\\
{C_{{M_{Ba{{(OH)}_2}}}}} = \dfrac{{0,25}}{{0,4}} = 0,625M\\
c)\\
BaS{O_3} + 2HCl \to BaC{l_2} + S{O_2} + {H_2}O\\
{n_{HCl}} = 2{n_{BaS{O_3}}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{30}} = 36,5g\\
{V_{HCl}} = \dfrac{{36,5}}{{1,15}} = 31,74ml
\end{array}\)