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H2O2 + 2KI overset("40% yield")to I2 + ...

`H_2O_2 + 2KI overset("40% yield")to I_2 + 2KOH`
` H_2O_2 + 2KMnO_4 + 3H_2SO_4 overset("50% yield ")to K_2SO_4 + 2MnSO_4 + 3O_2 + 4H_2O`
`150` ml of `H_2O_2` sample was divided into two parts. First part was treated with KI and Formed KOH required 200 ml. of `M//2 H_2SO_4` for neutralisation.Other part was trated with `KMnO_4` yielding 6.74 litre of `O_2` at STP.Using % yield indicated find volume stregth of `H_2O_2` sample used.

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