How many litres of `Cl_(2)` at STP will be liberated by the oxidation of `NaCl` with `10 g KMnO_(4)` in acidic medium: (Atomic weight: `Mn=55 and K=39)`

Options

A. `3.54 litres`
B. `7.08 litres`
C. `1.77 litres`
D. none of these

Answer

A

Explanation

`Cl^(-)+KmnO_(4) rarr Mn^(2+)+Cl_(2)`
Meq of `NaCl=` Meq of `KMnO_(4)`
`molexxn-factor=(10)/(158)xx(5)/(2)`
`molexx1=(10)/(158)xx(5)/(2)`
volume of `Cl_(2)=(10)/(158)xx(5)/(2)xx22.4=3.54 L`