Đáp án:

\( {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = 1,6M;\\{C_{M{\text{ NaOH}}}}= 0,8M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)

\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)

Ta có:

\({n_{KMn{O_4}}} = \dfrac{{50,56}}{{158}} = 0,32{\text{ mol}} \\\to {{\text{n}}_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}}} = 0,8{\text{ mol}}\)

\({n_{NaOH}} = 0,5.4 = 2{\text{ mol > 2}}{{\text{n}}_{C{l_2}}}\)

Vậy \(NaOH\) dư, dung dịch chứa \(NaCl;NaClO;NaOH\)

\({n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,8{\text{ mol;}}{{\text{n}}_{NaOH}} = 2 - 0,8.2 = 0,4{\text{ mol}}\)

\( \to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \dfrac{{0,8}}{{0,5}} = 1,6M;\\{C_{M{\text{ NaOH}}}} = \dfrac{{0,4}}{{0,5}} = 0,8M\)