Em tham khảo nha :
\(\begin{array}{l}
a)\\
{H_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2HCl\\
b)\\
{n_{{H_2}S{O_4}}} = 0,1 \times 2 = 0,2mol\\
{n_{BaC{l_2}}} = 0,1 \times 1 = 0,1mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,1}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1mol\\
{m_{BaS{O_4}}} = 0,1 \times 233 = 23,3g\\
c)\\
{n_{{H_2}S{O_4}d}} = 0,2 - 0,1 = 0,1mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,1}}{{0,1 + 0,1}} = 0,5M\\
{C_{{M_{HCl}}}} = \dfrac{{0,2}}{{0,2}} = 1M
\end{array}\)
