$\begin{array}{} &{n_{HCl}} = \frac{{21,3}}{{36,5}} = \frac{{213}}{{365}}(mol)\\ &{n_{NaOH}} = \frac{{20 \times 40\% }}{{40}} = 0,2(mol)\\ PTHH:\\ \quad \quad \quad HCl + NaOH \to NaCl + {H_2}O\\ 0,2\quad 0,2\quad \quad 0,2 \end{array}$ $Do\quad \frac{{213}}{{365}} > 0,2$ nên NaOH hết Dung dịch sau phản ứng sẽ gồm: $NaCl(0,2mol);\quad HCl(\frac{{28}}{{73}}mol)$
Ta có : ${m_{dung\;dich\;sau\;pu}} = 21,3 + 20 = 41,3g$
Nên $\% NaCl = \frac{{0,2 \times 58,5}}{{41,3}} = 28,32\% \\ \% HCl = \frac{{28 \times 36,5}}{{73 \times 41,3}} = 33,9\% $