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

In such type of problems, in order to simplify the calculation, let us assume that the intial weight of the mixture is `100 g`.
The the final weight of the mixture after heating in air will be `105 g`.
Let `x` be the weight of `FeO` in the initial mixture, then the weight of `Fe_(3)O_(4) = 100 - x`
When the mixture is heating in air `(O_(2))`:
`4 FeO + O_(2) rarr 2 Fe_(2) O_(3)`
`4 Fe_(3) O_(4) + O_(2) rarr 6 Fe_(2) O_(3)`
`implies 4 "moles" FeO -= 2 "moles of" Fe_(2)O_(3)`
`implies (x)/(72) "moles" -= (x)/(144) "moles of" Fe_(2) O_(3)`
`4 "moles of" Fe_(3) O_(4) -= 6 "moles of" Fe_(2_ O_(3)`
`implies (100 - x)/(232) "moles" -= (6)/(4) ((100 - x)/(232)) "moles" of Fe_(2) O_(3)`
Total moles of `Fe_(2)O_(3) = (x)/(144_ + (6)/(4) ((100 - x)/(232))`
Weight of `Fe_(2)O_(3) = [(x)/(144) + (6)/(4) ((100 - x)/(232))] 160 = 105`
Solving for `x`, we get,
`x = 20.25 g =` Weight of `FeO`
`implies "Weight of" Fe_(3)O_(4) = 100 - x = 79.75 g`
`% FeO = 20.25` and `% Fe_(3) O_(4) = 7.75`