Đáp án:
\(\begin{array}{l}
a,{m_{BaS{O_4}}} = 23,3g\\
b,C{\% _{HCl}} = 3,11\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
BaC{l_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2HCl\\
{m_{BaC{l_2}}} = \dfrac{{208 \times 10\% }}{{100\% }} = 20,8g\\
\to {n_{BaC{l_2}}} = 0,1mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 50g\\
\to {m_{{H_2}S{O_4}}} = \dfrac{{50 \times 19,6\% }}{{100\% }} = 9,8g\\
\to {n_{{H_2}S{O_4}}} = 0,1mol\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1mol\\
\to {m_{BaS{O_4}}} = 23,3g
\end{array}\)
\(\begin{array}{l}
b,\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
\to {m_{HCl}} = 7,3g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{\rm{BaS}}{O_4}}} = 234,7g\\
\to C{\% _{HCl}} = \dfrac{{7,3}}{{234,7}} \times 100\% = 3,11\%
\end{array}\)