Guest

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.

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. 

Grade:12th pass

1 Answers

Sakshi
askIITians Faculty 652 Points
8 years ago
2KMnO4 + H202 = 2KOH + MnO2 +O2
According to the above reaction, 2 moles of KMnO4 react with H2O2 = 1 mol
158*2g KMnO4 react with H2O2 = 34g
0.632 g KMnO4 react with H2O2 = 34*0.632/(158*2)
= 0.068g
But according to the given ques, mass of H2O2 used for the reaction = 0.4 g
thus, the percentage purity is 0.068*100/0.4 = 17%

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free