A mixture of FeO and Fe3O4 when heated in air to a constant mass gains 5% by mass.
calculate the composition of mixture
Answers
Answered by
88
Lets take weight of FeO be 'X'
& Fe3O4 be "y" gm
reactions occured 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
solving we get,
X= 21.06 & Y = 78.94
= 21.06 % & that of Fe3O4
= 78.94 %
Answered by
27
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 %
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 %
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