1.0g sample of `KCl_3` was heated under such condition that a part of it decomposed according to the equation.
`2KClO_3to2KCl+3O_2` and the remaining underwent change according to the equation.
`4KClO_3to3KClO_4+KCl`.
If the amount of `O_2` evolved was `145.8` mL at STP. Calculate the percentage by weight of `KClO_4` in the residue.

Explanation

`underset(245g)(2KClO_(3))tounderset(149g)(2KCl)+underset(96g)(3O_(2))`
`underset(490g)(4KClO_(3))tounderset(415.5g)(3KClO_(4))+underset(74.5g)(KCl)`
Let x g of `KClO_(3)` be converted to KCl and `O_(2)` and `(1-x)` g be converted to `KClO_(4)` and KCl.
Weight of `O_(2)=146.8xx(32)/(22400)=0.2097g`
`x=245xx(0.2097)/(96)=0.5352g`
Weight of `KClO_(4)` formed `=415.5xx((1-x))/(490)=0.3941g`
Weight of residue `=1-` weight of `O_(2)=1-0.2097g`
`therefore of KClO_(4)` in the residue`=(0.3941xx100)/((1-0.2097))=49.87`