A mixture of `FeO` and `Fe_(3)O_(4)` when heated in air to a constant weight, gains 5% of its weight. Find the composition of the intial mixutre.

Explanation

When FeO and `Fe_(3) O_(4)` are heated, both change to ` Fe_(2) O_(3)` . Let the weights of FeO and `Fe_(3) O_(4)` be x g and y respectively .
`:.` when FeO and `Fe_(3) O_(4)` change completely to `Fe_(2) O_(3)`
the wt. of `Fe_(2)O_(3) = (105)/( 100) xx ( x + y) = 1. 05 ( x + y) g`
Now,
`{:(FeO + Fe3 O_(4) to Fe_(2) O_(3)),("xg yg 1 . 05 (x + y) g "):}`
Applying POAC for Fe atoms,
moles of Fe in FeO + moles of Fe in `Fe_(3) O_(4)` = moles of Fe in `Fe_(2) O_(3)`
`1 xx` moles of FeO + `3 xx ` moles of `Fe_(3) O_(4) = 2 xx ` moles of `Fe_(2) O_(3)`
`(x)/(72) + (3y)/( 232) = (2 xx 1.05 (x + y))/( 160) [{:(" "FeO = 72),(Fe_(3)O_(4)=232),(Fe_(2)O_(3)=160):}]`
Dividing by y, we get
`(1)/( 72) xx (x)/( y) + (3)/( 232) = ( 2 xx 1.05) xx (x)/(y) + ( 2 + 1.05)/( 160)`
`(x)/(y) ((1)/( 27) - (2 . 1)/( 160)) = (2 . 1)/( 160) - (3)/( 232)`
`(x)/( y) = ( 81)/( 319) `
`:. % " of FeO" = (81)/( ( 81 + 310)) xx 100 = 20.02 %`
and `% of Fe_(3) O_(4) = (319)/(( 81 + 319)) xx 100 = 70 . 98 %`