A mixture of FeO and Fe3O4 when heated in air to constant weight gains 5% in its weight. The % of FeO in the sample is:
A
20.25%
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B
79.75%
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C
35.75%
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D
64.25%
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Solution
The correct option is A20.25% When FeO and Fe3O4 are heated both change to Fe2O3.
Let weight of FeO and Fe3O4 be xg and yg.
Total weight of reactant = (x+y)g
Since weight increases 5% on heating when FeO and Fe3O4 changes completely to Fe2O3.
Weight of Fe2O3=105100(x+y)=1.05(x+y) FeOxgm+Fe3O4ygm⟶2Fe2O31.05(x+y)
Applying POAC on Fe, x72+3y232=2×1.05(x+y)160 ∴xy=81319
% of FeO=8181+319×=20.25%
% of Fe3O4=79.75%