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H(2)O(2)+2KIoverset(40% "yield")rarrI(2)...

`H_(2)O_(2)+2KIoverset(40% "yield")rarrI_(2)+2KOH`
`H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)overset(50% "yield")rarr K_(2)SO_(4)+2MnSO_(4)+3O_(2)+4H_(2)O`
150mL of `H_(2)O_(2)` sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of `M//2H_(2)SO_(4)` for neutralisation. Other part was treated with `KMnO_(4)` yielding 6.74 litre of `O_(2)` at 1 atm. and 273 K. Using % yield indicated find volume strength of `H_(2)O_(2)` sample used.

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H_2O_2+2KI rarr 40% yield I_2 + 2KOH H_2O_2+2KMnO_4+3H_2SO_4 rarr 50% yield K_2SO_4+2KMnSO_4+3O_2 +4H_2O 150mL of H_2 O_2 sample was divided into two parts. First part was treated with KI and formed KOH required 200mL of M/2H_2SO_4 for neutralisation. Other part was treated with KMnO_4 yielding 6.74 litre of O_2 at 1 atm. and 273K . Using % yield indicated find volume strength of H_2O_2 sample used.

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.

KMnO_(4) + H_(2)SO_(4) + H_(2)C_(2)O_(4) rarr K_(2)SO_(4)+MnSO_(4) + CO_(2) + H_(2)O . Balance this equation.

(i) K_(2)Cr_(2)O_(7)+KI+H_(2)SO_(4)to K_(2)SO_(4)+Cr_(2)(SO_4)_3+I_(2)+H_(2)O .