Đáp án:

\(\begin{array}{l}
b)\\
{m_{NaCl}} = 5,85g\\
c)\\
{C_{{M_{NaCl}}}} = 0,2M\\
{C_{{M_{HCl}}}} = 0,4M\\
d)\\
{V_{KOH}} = 200ml
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
HCl + NaOH \to NaCl + {H_2}O\\
b)\\
{n_{HCl}} = {C_M} \times V = 0,3 \times 1 = 0,3mol\\
{n_{NaOH}} = {C_M} \times V = 0,2 \times 0,5 = 0,1mol\\
\dfrac{{0,3}}{1} > \dfrac{{0,1}}{1} \Rightarrow HCl\text{ dư}\\
{n_{NaCl}} = {n_{NaOH}} = 0,1mol\\
{m_{NaCl}} = n \times M = 0,1 \times 58,5 = 5,85g\\
c)\\
{n_{HC{l_d}}} = {n_{HCl}} - {n_{NaOH}} = 0,2mol\\
{C_{{M_{NaCl}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,3 + 0,2}} = 0,2M\\
{C_{{M_{HCl}}}} = \dfrac{n}{V} = \dfrac{{0,2}}{{0,3 + 0,2}} = 0,4M\\
d)\\
HCl + KOH \to KCl + {H_2}O\\
{n_{KOH}} = {n_{HC{l_d}}} = 0,2mol\\
{V_{KOH}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,2}}{1} = 0,2l = 200ml
\end{array}\)