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How many litres of Cl(2) at STP will be ...

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)`

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The correct Answer is:
3.54
meq of `Cl_(2) = m eq KMnO_(4)`
`W/71 xx 2 xx 1000 = 10/158 xx 1000 xx 5`
w = 11.23g `V_(Cl_(2)) = 22.4 xx 11.23 /71 = 3.54 l`
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