Đáp án:
\(\begin{array}{l}
{V_{{\rm{dd}}Ba{{(OH)}_2}}} = 1,05l\\
{m_{BaS{O_3}}} = 21,7g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
S{O_2} + Ba{(OH)_2} \to BaS{O_3} + {H_2}O\\
{n_{S{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Ba{{(OH)}_2}}} = {n_{S{O_2}}} = 0,1\,mol\\
{n_{Ba{{(OH)}_2}}} = 0,1 \times 105\% = 0,105\,mol\\
{V_{{\rm{dd}}Ba{{(OH)}_2}}} = \dfrac{{0,105}}{{0,1}} = 1,05l\\
{n_{BaS{O_3}}} = {n_{S{O_2}}} = 0,1\,mol\\
{m_{BaS{O_3}}} = 0,1 \times 217 = 21,7g
\end{array}\)