Giải thích các bước giải:
\(\begin{array}{l}
5.\\
KOH + HCl \to KCl + {H_2}O\\
{n_{KOH}} = 0,2mol\\
{m_{HCl}} = \dfrac{{100 \times 36,5\% }}{{100\% }} = 36,5g\\
\to {n_{HCl}} = 1mol\\
\to {n_{HCl}} > {n_{KOH}} \to {n_{HCl}}dư\\
\to {n_{HCl(pt)}} = {n_{KOH}} = 0,2mol\\
\to {n_{HCl}}dư = 0,8mol \to {m_{HCl}}dư = 29,2g\\
{n_{KCl}} = {n_{KOH}} = 0,2mol \to {m_{KCl}} = 14,9g\\
6.\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{100\% }} = 39,2g\\
\to {n_{{H_2}S{O_4}}} = 0,4mol\\
\to {n_{{H_2}S{O_4}}} > {n_{Ba{{(OH)}_2}}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{{\rm{BaS}}{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,1mol \to {m_{{\rm{BaS}}{O_4}}} = 23,3g
\end{array}\)
