1 g sample of `KClO_(3)` was heated under such conditions that a part of it decomposed according to the equation :
`2KClO_(3) rarr 2KCl + 3O_(2)` and the remaining underwent change according to the equation : `4KClO_(3)rarr 3KClO_(4)+KCl`
If the amount of `O_(2)` evolved was 146.8 ml at S.T.P., calculate the % by weight of `KClO_(4)` in the reside.

Explanation

`KCIO_(3) to KCI + O_(2)`
Applying POAC for O atoms in the eqn. (i),
,oles of O in `KCIO_(3) = ` moles of O in `O_(2)`
`3 xx` moles of `KCIO_(3) = 2 xx ` moles of ` O_(2)`
`3 xx (wt . if KCIO_(3))/( "mol. wt. of " KCIO_(3)) = 2 xx ("volume at NTP (mL))/( 22400)`
Wt. of `KCIO_(3) = (2 xx 146 . 8 xx 122.5)/( 3 xx 22400)`
= 0.5358 g
Again applying POAC for K atoms,
moles of K atoms in `KCIO_(3)` = moles of K atoms in KCI
`("wt. of " KCIO_(3))/( "mol. wt. of " KCIO_(3)) = ("wt. of KCI")/( "mol. wt. of KCI")`
Wt. of KCI ` = (0. 5358)/(122.5) xx 74.5 = 0.3260g" ". . . (i)`
In the second reaction :
the amount of `KCIO_(3)` left = 1 - 0.5358 = 0.4642 g
We have,
`KCIO_(3) to KCIO_(4) + KCI`
0.4642 g.
Applying POAC for O atoms,
moles of O in `KCIO_(3)` = moles of O in ` KCIO_(4)`
`3 xx ` moles of `KCIO_(3) = 4 xx ` moles of `KCIO_(4)`
`3 xx ("wt. of " KCIO_(3))/("mol. wt. of "KCIO_(3))= 4 xx ("wt. of " KCIO_(4))/( "mol. wt. of " KCIO_(4))`
Wt. of `KCIO_(4) = (3 xx 0.4642 xx 138.5)/( 122.5 xx 4)`
= 0.3937 g . . . (ii)
WT. of KCI produced by second reaction
= wt. of `KCIO_(3)` - wt. of `KCIO_4)`
= 0.4642 - 0.3937 = 0.0705 g . . . (iii)
Now since on heating `KCIO_(3),O_(2)` shall escape out, the substance as residue are KCI produced by the reaction (i) and (ii) and `KCIO_(4)`
Wt. of residue = (i) + (ii) + (iii)
= 0.3260 +0.3937 +0.0705
= 0.7902 g
`:. % of KCIO_(4)` in the residue `= (0.3937)/(0.7902) xx 100`
`= 49.8%`