Giải thích các bước giải:

\(\begin{array}{l}
a)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = 0,1mol\\
{n_{HCl}} = 0,2mol\\
 \to {n_{HCl}} > {n_{NaOH}} \to {n_{HCl}}dư\\
 \to {n_{HCl(pt)}} = {n_{NaCl}} = {n_{NaOH}} = 0,1mol\\
 \to {n_{HCl(dư)}} = 0,1mol\\
 \to C{M_{HCl(dư)}} = C{M_{NaCl}} = \dfrac{{0,1}}{{0,1 + 0,1}} = 0,5M
\end{array}\)

\(\begin{array}{l}
b)\\
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{n_{HCl}} = 0,1mol\\
 \to {n_{BaC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,05mol\\
 \to C{M_{BaC{l_2}}} = \dfrac{{0,05}}{{0,1 + 0,1}} = 0,25M\\
c)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,219mol\\
 \to {n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,438mol\\
 \to {V_{KOH}} = \dfrac{{0,438}}{{1,5}} = 0,292l\\
d)\\
NaOH + HCl \to NaCl + {H_2}O\\
{m_{NaOH}} = \dfrac{{200 \times 10\% }}{{100\% }} = 20g\\
 \to {n_{NaOH}} = 0,5mol\\
 \to {n_{HCl}} = {n_{NaOH}} = 0,5mol\\
 \to {m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{18,25}}{{3,65\% }} \times 100\%  = 500g\\
e)\\
KOH + HCl \to KCl + {H_2}O\\
{n_{KOH}} = 0,1mol\\
 \to {n_{HCl}} = {n_{KOH}} = 0,1mol\\
 \to {m_{HCl}} = 0,1 \times 36,5 = 3,65g\\
 \to C{\% _{HCl}} = \dfrac{{3,65}}{{200}} \times 100\%  = 1,285\% 
\end{array}\)