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

The chemical equations are:
`2FeO + 1/2 O_(2) to Fe_(2)O_(3)`……(i)
`2FeO + 1/2O_(2) to Fe_(2)O_(3)`……..(i)
`underset(462.8)(2Fe_(2)O_(3) + 1/2O_(2)) to underset(478.8)(3Fe_(2)O_(3))`……..(ii)
Let the weight of mixture = 100g
weight of FeO = xg
Weight of `Fe_(3)O_(4) = (100-x)g`
According to equation (i),
143.6 g of FeO produce `Fe_(2)O_(3) = 159.6`g
xg of FeO produce `Fe_(2)O_(3) = 159.6/143.6 xx x`
According to equation (ii)
462.8 g of `Fe_(3)O_(4)` produce `Fe_(2)O_(3) = 478.8`g
(100-x) g of `Fe_(3)O_(4)` produce `Fe_(2)O_(3) = = 478.8/462.8 xx (100-x)`
Total mass of `Fe_(2)O_(3)` formed by heating
`(159.6)/(143.6)x + (478.8)/(462.8) (100-x) = 105`
Solving, we get
`therefore` Mass of FeO = 19.92
Mass of `Fe_(3)O_(4) = 100-19.92 = 80.08`
Mass % FeO `=(19.92)/100 xx 100 19.92`
Mass % `Fe_(3)O_(4) = 80.08`.