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

Let the total mass of mixture = 100 g
Let the mass of FeO in the mixture = x g
`:.` Mass of `Fe_(3)O_(4)` in the mixture = (100-x)g
Both the oxides on heating in air are converted to `Fe_(2)O_(3)`
Step I. Calculation of mass of `Fe_(2)O_(3)` from `FeO`
`{:(2FeO+,(1)/(2)O_(2)rarr,Fe_(2)O_(3)),(2(56+16),,2xx56+3xx16),(=144g,,=160g):}`
144g of FeO upon oxidation form `Fe_(2)O_(3)=160g`
`:.` xg of FeO upon oxidation form `Fe_(2)O_(3)=(160)/(144)xx xg`
Step II. Calculation of mass of `Fe_(2)O_(3)` from `Fe_(3)O_(4)`
`{:(2Fe_(3)O_(4)+(1)/(2)O_(2),overset("Heat")(rarr),3Fe_(2)O_(3)),(2(3xx56+4xx16),,3(2xx56+3xx16)),(=464g,,=480g):}`
464g of `Fe_(3)O_(4)` upon oxidation form `Fe_(2)O_(3) = 48` g
(100-x)g of `Fe_(3)O_(4)` upon oxidation form `Fe_(2)O_(3) = (480)/(464)(100-x)g`
Step III. Calculation of percentage composition of the mixture
Total mass of `Fe_(2)O_(3)` actually formed `= 100 +5 = 105` g
`:. (160x)/(144)+(480)/(464)(100-x)=105,1.11x+1.034(100-x)=105`
`1.11x+103.4-1.034x =105`
`1.11x-1.034x=105-103.4 or0.076=1.6orx=(1.6)/(0.076)=21g`
% of FeO in the sample `= ((21 g))/((100 g))xx100 = 21 %`
% of `Fe_(3)O_(4)` in the sample `= 100 - 21 = 79 %`.