Đáp án:

\({C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = 0,3M;\;{C_{M{\text{ NaOH dư}}}} = 0,4M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)

Ta có: \({n_{Mn{O_2}}} = \frac{{13,05}}{{55 + 16.2}} = 0,15{\text{ mol}} \to {{\text{n}}_{C{l_2}}} = {n_{Mn{O_2}}} = 0,15{\text{ mol}}\)

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

Vì \({n_{NaOH}} = 0,5.1 = 0,5{\text{  > 2}}{{\text{n}}_{C{l_2}}}\) nên NaOH dư.

\(\to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,15{\text{ mol; }}{{\text{n}}_{NaOH{\text{ dư}}}} = 0,5 - 0,15.2 = 0,2{\text{ mol}}\)

\(\to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \frac{{0,15}}{{0,5}} = 0,3M;\;{C_{M{\text{ NaOH dư}}}} = \frac{{0,2}}{{0,5}} = 0,4M\)