\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
PT: \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
Theo PT: \(n_{Cl_2}=\dfrac{5}{2}n_{KMnO_4}=0,5\left(mol\right)\)
\(n_{NaOH}=0,5.4=2\left(mol\right)\)
PT: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{2}{2}\), ta được NaOH dư.
Theo PT: \(\left\{{}\begin{matrix}n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,5\left(mol\right)\\n_{NaOH\left(pư\right)}=2n_{Cl_2}=1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{NaOH\left(dư\right)}=2-1=1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,5}{0,5}=1\left(M\right)\\C_{M_{NaOH}}=\dfrac{1}{0,5}=2\left(M\right)\end{matrix}\right.\)