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A mixture of FeO and Fe(3)O(4) when heat...

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.

Text Solution

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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`
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