Đáp án: a. \({{\text{m}}_{_{C{l_2}}}}{\text{  =  2,13 gam}}\)

 b. \({{\text{C}}_{M{\text{ NaOH dư}}}}{\text{   =  3,97 M}}\)

\({{\text{C}}_{M{\text{ NaCl}}}}{\text{   =  0,03 M}}\)

\({{\text{C}}_{M{\text{ NaClO}}}}{\text{  =   0,03 M}}\)

Giải thích các bước giải:

 a. \(2KMn{O_4}{\text{ }} + {\text{ }}16HCl{\text{ }} \to {\text{ }}2KCl{\text{ }} + {\text{ }}MnC{l_2}{\text{ }} + {\text{ }}5C{l_2}{\text{ }} + {\text{ }}8{H_2}O\)

\({n_{KMn{O_4}}} = \frac{{1,896}}{{158}} = 0,012{\text{ mol}}\)

\({\text{Theo PTHH: }}{{\text{n}}_{C{l_2}}} = \frac{5}{2}{n_{KMn{O_4}}} = \frac{5}{2}{\text{ }}x{\text{ }}0,012 = {\text{ }}0,03{\text{ mol}}\)

\({{\text{m}}_{_{C{l_2}}}} = {\text{ 0,03 x 71  =  2,13 gam}}\)

b. \(C{l_2} + {\text{ 2NaOH }} \to {\text{ NaCl  +  NaClO  +  H2O}}\)

\({{\text{n}}_{NaOH}}{\text{  =  1 x 4  =  4 mol}}\)

\({\text{Do }}\frac{{{{\text{n}}_{NaOH}}}}{2}{\text{  >  }}\frac{{{{\text{n}}_{C{l_2}}}}}{1}{\text{ }} \to {\text{ C}}{{\text{l}}_2}\) hết

\(TheoPTHH:{\text{ }}{{\text{n}}_{NaCl}}{\text{  =  }}{{\text{n}}_{NaClO}}{\text{  =  }}{{\text{n}}_{NaOH{\text{ pứ}}}}{\text{  =  }}{{\text{n}}_{C{l_2}}}{\text{  =  0,03 mol}}\)

\({n_{NaOH{\text{ dư}}}}{\text{  =  }}{{\text{n}}_{NaOH{\text{ ban đầu}}}} - {n_{NaOH{\text{ pứ }}}} = {\text{ 4  -  0,03 x 2  =  3,94 mol}}\)

Dung dịch sau phản ứng: NaOH dư, NaCl, NaClO

\({{\text{C}}_{M{\text{ NaOH dư}}}}{\text{  =  }}\frac{{{{\text{n}}_{NaOH{\text{ du}}}}}}{V}{\text{  =  }}\frac{{3,94}}{1}{\text{  =  3,94 M}}\)

\({{\text{C}}_{M{\text{ NaCl}}}}{\text{  =  }}\frac{{{{\text{n}}_{NaCl}}}}{V}{\text{  =  }}\frac{{0,03}}{1}{\text{  =  0,03 M}}\)

\({{\text{C}}_{M{\text{ NaClO}}}}{\text{  =  }}\frac{{{{\text{n}}_{NaClO}}}}{V}{\text{  =  }}\frac{{0,03}}{1}{\text{  =  0,03 M}}\)

Em tham khảo nha:

\(\begin{array}{l}
a)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{KMn{O_4}}} = \dfrac{{1,896}}{{158}} = 0,012\,mol\\
{n_{C{l_2}}} = 0,012 \times \frac{5}{2} = 0,03\,mol\\
{m_{C{l_2}}} = 0,03 \times 71 = 2,13g\\
b)\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
{n_{NaOH}} = 4 \times 1 = 4\,mol\\
{n_{NaOH}}  \text{ dư }= 4 - 0,03 \times 2 = 3,94\,mol\\
{n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,03\,mol\\
{C_M}NaOH \text{ dư } = \dfrac{{3,94}}{1} = 3,94M\\
{C_M}NaCl = {C_M}NaClO = \dfrac{{0,03}}{1} = 0,03M
\end{array}\)