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. (a)`3.54`
B. (b)`7.08`
C. (c )`1.77`
D. (d)None of these

Answer

A

Explanation

`Cl^(-)+MnO_(4)^(-) rarr Mn^(+2)+Cl_(2)`
`"Eq of" Cl_(2)="Eq of" KMnO_(4)`
`2 ["mole of" Cl_(2)]=5[10/158]`
mole of `Cl_(2) =50/(2xx158)=0.15823` mole
volume of `Cl_(2)` at `STP=0.15823xx22.4=3.54 L`