Đáp án:

Bạn tham khảo lời giải ở dưới nhé ! 

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
A,\\
2KMn{O_4} + 16HCl \to MnC{l_2} + 2KCl + 5C{l_2} + 8{H_2}O\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O
\end{array}\)

\(\begin{array}{l}
B,\\
{n_{KMn{O_4}}} = 0,2mol\\
 \to {n_{MnC{l_2}}} = \dfrac{1}{2}{n_{KMn{O_4}}} = 0,1mol\\
 \to {n_{KCl}} = {n_{KMn{O_4}}} = 0,2mol\\
 \to {n_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}}} = 0,5mol\\
{n_{NaOH}} = 1,6mol\\
 \to {n_{NaOH}} > {n_{C{l_2}}} \to {n_{NaOH}}dư
\end{array}\)

\(\begin{array}{l}
 \to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,5mol\\
 \to {n_{NaOH}} = 2{n_{C{l_2}}} = 1mol\\
 \to {n_{NaOH(dư)}} = 0,6mol\\
 \to C{M_{MnC{l_2}}} = \dfrac{{0,1}}{{0,4}} = 0,25M\\
 \to C{M_{KCl}} = \dfrac{{0,2}}{{0,4}} = 0,5M\\
 \to C{M_{NaCl}} = C{M_{NaClO}} = \dfrac{{0,5}}{{0,4}} = 1,25M\\
 \to C{M_{NaOH(dư)}} = \dfrac{{0,6}}{{0,4}} = 1,5M
\end{array}\)

\(\begin{array}{l}
C,\\
{m_{NaCl}} = 29,25g\\
{m_{NaClO}} = 37,25g
\end{array}\)

\(\begin{array}{l}
2.\\
A,\\
2KMn{O_4} + 16HCl \to MnC{l_2} + 2KCl + 5C{l_2} + 8{H_2}O\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O
\end{array}\)

\(\begin{array}{l}
B,\\
{n_{KMn{O_4}}} = 0,04mol\\
 \to {n_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}}} = 0,1mol\\
 \to {V_{C{l_2}}} = 2,24l
\end{array}\)

câu b tính a là gì vậy bạn??