0.261 g of a sample of pyrolusite was heated with excess of HCl and the chlorine evolved was passed in a solution of KI. The liberated iodine required 90 mL `(N)/(30) Na_(2)S_(2)O_(3)`. Calculate the percentage of `MnO_(2)` in the sample.
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`MnO_(2)+4HCl to MnCl_(2)+2H_(2)O+Cl_(2)` `2KI+Cl_(2) to 2KCl+I_(2)` `2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI` `90 mL (N)/(30) Na_(2)S_(2)O_(3)-=90 mL (N)/(30)I_(2)` `-=90 mL (N)/(30)Cl_(2)` `-=90 mL(N)/(30)MnO_(2)` Eq. mass of `MnO_(2)=("Mol. mass")/(2)=(87)/(2)` [Since, change in O.N. is from 4 to 2] Amount of `MnO_(2)=(87)/(2xx30)xx(90)/(1000)=0.1305 g` % of `MnO_(2)=(0.1305)/(0.261)xx100=50`
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