10 cc of `H_(2) O_(2)` solution when reacted with KI solution produced 0 . 5 g of iodine . Calculate the percentage purity of `H_(2) O_(2)`

Explanation

`H_(2) O_(2) + KI to KOH + I_(2)`
First, applying POAC for H atoms to calculate moles of KOH, then applying POAC for K atoms to calculate moles of KI, and then finally applying POAC for I atoms to calculate moles of iodine.
POAC for H atoms :
moles of H in `H_(2)O_(2) ` = moles of H in KOH
`2 xx ` moles of `H_(2)O_(2) = 1 xx ` moles of KOH . . . (i)
Applying POAC for K atons,
moles of K atoms in KI = moles o f K atoms in KOH
`1 xx` moles of KI = ` 1 xx ` moles of KOH
`=2 xx ` moles of `H_(2) O_(2)`
[from the equ. (i)]
Applying POAC for I atoms,
moles of I atoms in KI = moles of I atoms in` I_(2)`
or moles of `I_(2) = (1)/(2) xx ` moles of KI
`= (1)/(2) xx 2 ` moles of `H_(2) O_(2)` [from the eqn. (ii)]
Now, `(wt. of I_(2))/( mol. wt. of I_(2)) = (wt. of H_(2) O_(2))/(mol . wt. of H_(2)O_(2))`
Suppose x is the wt. of `H_(2)O_(2)` . Then,
`(0. 5)/( 254) = (x)/( 34)`
`x = (34 xx 0 . 5)/( 254) = 0.0669 g` .
`% of H_(2)O_(2) = (0.0669)/(10) xx 100 = 0 . 669 %`