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A 0.5 g sample containing MnO_(2) is treated with HCl liberating Cl_(2). The Cl_(2) is passed into a solution of KI and 30.0 cm^(3) of 0.1 M Na_(2)S_(2)O_(3) are required to tirate the liberated iodine. Calculate the percentage of MnO_(2) in the sample. |
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Answer» `-=30.0 mL 0.1 N I_(2)` `-=30.0 mL 0.1 N Cl_(2)` `-=30.0 mL 0.1 N MnO_(2)` Amount of `MnO_(2)` present `=(NxxExxV)/(1000)` `=(1)/(10)xx(87)/(2)xx(30)/(1000)` `% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1` |
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