$n_{KMnO_4}=\dfrac{31,6}{158}=0,2(mol)$
$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O$
$\Rightarrow n_{Cl_2}=\dfrac{5}{2}n_{KMnO_4}=0,5(mol)$
$n_{NaOH}=\dfrac{145,8.20\%}{40}=0,729(mol)$
$2NaOH+Cl_2\to NaCl+NaClO+H_2O$
$\Rightarrow Cl_2$ dư
$n_{NaCl}=n_{Cl_2\text{pứ}}=\dfrac{0,729}{2}=0,3645(mol)$
Clo được hấp thụ hết nên lượng clo dư tan hết vào dd.
$m_X=0,5.71+145,8=181,3g$
$\to C\%_{NaCl}=\dfrac{0,3645.58,5.100}{181,3}=11,76\%$