Em tham khảo nha :
\(\begin{array}{l}
a)\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
b)\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 19,6}}{{100}} = 19,6g\\
{n_{{H_2}S{O_4}}} = \dfrac{{19,6}}{{98}} = 0,2mol\\
{n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{BaS{O_4}}} = 0,2 \times 233 = 46,6g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}S{O_4}}} = 0,4mol\\
{m_{HCl}} = 0,4 \times 36,5 = 14,6g\\
{m_{{\rm{dd}}spu}} = 500 + 100 - 46,6 = 553,4g\\
C{\% _{HCl}} = \dfrac{{14,6}}{{553,4}} \times 100\% = 2,64\%
\end{array}\)
