Question

a mixture of FeO and Fe3O4 when heated in air to a constant mass gains by 5 % by mass.
calculate the composition of mixture

Answer 1

 Lets take weight  of FeO be 'X' 

& Fe3O4 be "y" gm

reactions occurring as follows: 

2 FeO + 1/2 O2 ----> Fe2O3 
Fe3O4 + 1/2 O2 ------> 3Fe2O3 

Molar mass of FeO =144 g 

and Fe2O3 =160 gms. 

144 g  of FeO gives to  160 g Fe2O3 

'X' g FeO will give up  160 x X/144 gm Fe2O3                                        ----(1) 
weight of Fe2O3 created by  Y gm Fe3O4  = 160 x 3Y/464                    ----(2) 

considering the total weight i.e. (X+Y) = 100.                                            -(3) 

 from equation  1,2 & 3 . 

(160 x X /144) +160 x 3y/464 = 105 

by solving we get, 

X= 21.06 & Y = 78.94 
= 21.06 % & that of Fe3O4 

= 78.94 %

Answer 2

The chemical equations are 

2Fe + 1/2 O2 -----> Fe2O3 -----(1) 

2(55.8 + 16)            2 x 55.6 + 3 X 16 
                                
= 143.6g                = 159 .6 g


2Fe3O4 + 1/2O2   -----> 3FeO3 ----(2) 

2(55.6 x 3 + 4 x 16)         3 (2 x 55.6 + 3 x 16) 

= 462.8 g                         = 478.8 g 


Let the  weight of mixture = 100 g 

weight of FeO = x g 

Weight of Fe3O4 = (100 - x) g 

ATQ (from ( 1)

143.6 g of FeO produce Fe2O3 = 159.6 g 

x g of FeO produce Fe2O3 = (159.6/143.6 )x

ATQ (from (2) 

462.8 g of Fe3O4 produce Fe2O3 = 478.8 g 

(100 - x )g of Fe3O4 produce Fe2O3 = (478.8/462.8) (100 - x) 

Total mass of Fe2O3 Produce by heating 

(159.6/143.6) x + (478.8/462.8) (100 - x) 

Gain in weight of mixture = 100 x 5 / 100 = 5 g 

Total Fe2O3 formed = 100 + 5 = 105 g 

(159.6/143.6)x + (478.8/462.8)(100 -x) = 105 

By solving we get 

x = 19.92 

mass of FeO = 19.92 

Mass of Fe3O4 = 100 - 19.92 = 80 .08 

Mass % of FeO = 19.92 /100 x 100 = 19.92 % 

Mass of Fe3O4 = 80.08 %