\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{182,5.40\%}{36,5}=2\left(mol\right)\)
PTHH: 2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{2}{16}\) => KMnO4 hết, HCl dư
PTHH: 2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,2-------------------------------------->0,5
=> \(V_{Cl_2}=0,5.22,4=11,2\left(l\right)\)