Đáp án:
\({m_{Ba{{(OH)}_2}}} = 171g\)
\(C\% BaC{l_2} = \dfrac{{166,4}}{{524,4}} \times 100\% = 31,73\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{HCl}} = \dfrac{{200 \times 7,3}}{{100 \times 36,5}} = 0,4mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 9,8}}{{100 \times 98}} = 0,2mol\\
\to {n_{Ba{{(OH)}_2}}} = 2{n_{HCl}} + {n_{{H_2}S{O_4}}} = 1mol\\
\to {m_{Ba{{(OH)}_2}}} = 171g
\end{array}\)
Dung dịch sau phản ứng có \(BaC{l_2}\)
\(\begin{array}{l}
{n_{BaC{l_2}}} = 2{n_{HCl}} = 0,8mol\\
\to {m_{BaC{l_2}}} = 166,4g\\
{n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
\to {m_{BaS{O_4}}} = 46,6g\\
{m_{dd}} = 171 + 200 + 200 - 46,6 = 524,4g\\
\to C\% BaC{l_2} = \dfrac{{166,4}}{{524,4}} \times 100\% = 31,73\%
\end{array}\)
