50g of a good sample of CaOCl2 is made to reac with CO2 the volume of Cl2 liberated at S.T.P

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CaOCl2 + CO2 -----> CaCO3 + Cl2
Accoding to stoichiometry of the reaction, 1 mole of CaOCl2 gives 1 mole pf Cl2
Thus,
Moles of CaOCl2 in ques = mass/MM = 50/198 = 0.25 moles
0.25 moles of CaOCl2 gives 0.25 moles of Cl2
1 mole of Cl2 at STP contains = 22.4 litre
0.25 mole of Cl2 at STP contains = 22.4*0.25 = 5.6 litre