Đáp án:

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Giải thích các bước giải:

Bài 9:

a,

\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
{n_{Mn{O_2}}} = 0,2mol\\
 \to {n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,2mol\\
{n_{NaOH}} = \dfrac{{145,8 \times 20}}{{100 \times 40}} = 0,729mol\\
 \to \dfrac{{{n_{NaOH}}}}{2} > {n_{C{l_2}}}
\end{array}\)

Suy ra NaOH dư

\(\begin{array}{l}
 \to {n_{NaOH}}dư= 0,729 - 2 \times 0,2 = 0,329mol\\
 \to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,2mol\\
{m_{{\rm{dd}}}} = 145,8 + 0,2 \times 71 = 160g\\
 \to C{\% _{NaOH}}dư= \dfrac{{0,329 \times 40}}{{160}} \times 100\%  = 8,225\% \\
 \to C{\% _{NaCl}} = \dfrac{{0,2 \times 58,5}}{{160}} \times 100\%  = 7,3125\% \\
 \to C{\% _{NaClO}} = \dfrac{{0,2 \times 74,5}}{{160}} \times 100\%  = 9,3125\% 
\end{array}\)

b,

\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
2KOH + C{l_2} \to KCl + KClO + {H_2}O\\
{n_{HCl}} = \dfrac{{20 \times 36,5}}{{100 \times 36,5}} = 0,2mol\\
 \to {n_{C{l_2}}} = \dfrac{1}{4}{n_{HCl}} = 0,05mol\\
{n_{KOH}} = 1mol\\
 \to \dfrac{{{n_{KOH}}}}{2} > {n_{C{l_2}}}
\end{array}\)

Suy ra KOH dư

\(\begin{array}{l}
 \to {n_{KOH}}dư= 1 - 2 \times 0,05 = 0,9mol\\
 \to {n_{KCl}} = {n_{KClO}} = {n_{C{l_2}}} = 0,05mol\\
 \to C{M_{KOH}}dư= \dfrac{{0,9}}{{0,5}} = 1,8M\\
 \to C{M_{KCl}} = C{M_{KClO}} = \dfrac{{0,05}}{{0,5}} = 0,1M
\end{array}\)