Đáp án:

\(\begin{array}{l}
{m_{{\rm{BaS}}{O_4}}} = 46,6g\\
C{\% _{BaC{l_2}}} = 10,4\% \\
C{\% _{HCl}} = 3,23\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2HCl\\
{m_{{H_2}S{O_4}}} = \dfrac{{98 \times 20\% }}{{100\% }} = 19,6g\\
 \to {n_{{H_2}S{O_4}}} = 0,2mol\\
 \to {n_{{\rm{BaS}}{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
 \to {m_{{\rm{BaS}}{O_4}}} = 46,6g\\
{n_{BaC{l_2}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
 \to {m_{BaC{l_2}}} = 41,6g\\
 \to C{\% _{BaC{l_2}}} = \dfrac{{41,6}}{{400}} \times 100\%  = 10,4\% \\
{n_{HCl}} = 2{n_{{H_2}S{O_4}}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}sauphảnứng}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{\rm{BaS}}{O_4}}} = 451,4g\\
 \to C{\% _{HCl}} = \dfrac{{14,6}}{{451,4}} \times 100\%  = 3,23\% 
\end{array}\)