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a solution of a 0.4g sample of H2O2 reacted with 0.632g of KMnO4 in the precence of sulphuric acid.calculate the percentage purity of the sample of H2O2.
2KMnO4 + H202 = 2KOH + MnO2 +O2According to the above reaction, 2 moles of KMnO4 react with H2O2 = 1 mol158*2g KMnO4 react with H2O2 = 34g0.632 g KMnO4 react with H2O2 = 34*0.632/(158*2)= 0.068gBut according to the given ques, mass of H2O2 used for the reaction = 0.4 gthus, the percentage purity is 0.068*100/0.4 = 17%
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