Đáp án:
\({m_{{\rm{BaS}}{O_4}}} = 23,3g\)
\(\begin{array}{l}
C{\% _{HCl}} = 1,5\% \\
C{\% _{{H_2}S{O_4}(dư)}} = 2,6\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2HCl\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 114g\\
\to {m_{{H_2}S{O_4}}} = \frac{{114 \times 20\% }}{{100\% }} = 22,8g \to {n_{{H_2}S{O_4}}} = 0,23mol\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2\% }}{{100\% }} = 20,8g \to {n_{BaC{l_2}}} = 0,1mol\\
\to {n_{{H_2}S{O_4}}} > {n_{BaC{l_2}}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{{H_2}S{O_4}(pt)}} = {n_{BaC{l_2}}} = 0,1mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,13mol\\
\to {m_{{H_2}S{O_4}(dư)}} = 12,74g\\
\to {n_{{\rm{BaS}}{O_4}}} = {n_{BaC{l_2}}} = 0,1mol \to {m_{{\rm{BaS}}{O_4}}} = 23,3g\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol \to {m_{HCl}} = 7,3g\\
{m_{{\rm{dd}}B}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{\rm{BaS}}{O_4}}} = 490,7g\\
\to C{\% _{HCl}} = \dfrac{{7,3}}{{490,7}} \times 100\% = 1,5\% \\
\to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{12,74}}{{490,7}} \times 100\% = 2,6\%
\end{array}\)
