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  4. How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO4 in acidic medium: (Atomic weight: Mn = 55 and K = 39)

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Q. 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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Solution:

$Cl^{-} + MnO_4^{-} \to Mn^{+2} + Cl_2$
$Eq\, of\, Cl_2 = Eq \,of\, KMnO_4$
$2\left[mole of CI_{2}\right]=5\left[\frac{10}{158}\right]$
mole of $Cl_2 =\frac{50}{2\times158}$=0.15823 mole
volume of $Cl_2$ at $STP = 0.15823 \times 22.4 = 3.54L$