What is the equivalent mass of HCl in the given reaction?
2KMnO4+16 HCl→2KCl+2MnCl2+5Cl2+8H2O
As only 16 moles of Cl− ions undergoes oxidation,(5Cl2 is product formed)
n - factor of HCl will be =1016=58 (N-factor of an acid is its basicity, in 16 moles of HCl, 10 moles of HCl is consumed)
Equivalent weight of HCl =Molar Massn - factor
Equivalent weight of HCl =36.5(58)=8×36.55=58.4