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1 gram of pyrolusite (MnO(2)) was boiled...

1 gram of pyrolusite `(MnO_(2))` was boiled with excess of concentrated `HCl` and the issuing gas was passed through a solution of potassium iodide when 1.27 g of iodide were liberated. What is the percentage of pure `MnO_(2)` in the sample ?

Text Solution

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Calculation of weight of pure `MnO_(2)`
The following partial equations are involved in the problem
`{:(" "MnO_(2)+4HClrarrMnCl_(2)+Cl_(2)+2H_(2)O),(" "2KI+Cl_(2)rarr2KCl+I_(2)),(bar("Add :"" "underset(55+32(=87g))(MnO_(2))+4HCl+2KIrarrMnCl_(2)+2KCl+2H_(2)underset(2xx127(=254g))(O+I_(2)))):}`
254g of `I_(2)` are obtained from `MnO_(2)=87` g
1.27 g of `I_(2)` are obtained from `MnO_(2)=(87)/(254)xx1.27 g = 0.435 g`
Step II. Calculation of percentage purity of the sample
Weight of pure `MnO_(2)=0.435 g`
Weight of the sample `= 1.0 g`
Percentage purity of the sample `= ("Weight of pure MnO"_(2))/("Weight of sample")=((0.435g))/((1.0g))xx100=43.5%`.
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