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Answer 1

Answer:

Let us consider weight of FeO & Fe3O4 be 'a' & 'b' gm respectively.

Consider the following reactions. 2 FeO + 1/2 O2 ------> Fe2O3

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

Molar mass of FeO =144 g and that of Fe2O3 is 160 gms.

Thus 144 g FeO gives 160 g Fe2O3

'a' g FeO will give 160 x a/144 gm Fe2O3 --------------(1)

Similarly, weight of Fe2O3 formed by ‘b’gm Fe3O4 = 160 x 3b/464 --------(2)

Assuming total weight i.e. (a+b) = 100. -------- (3)

As there is an increase of 5 % we get, from eqn. 1,2 & 3

. (160 x a /144) +160 x 3b/464 = 105

solving we get,

a= 21.06 & b = 78.94

Thus % of FeO = 21.06 % & that of Fe3O4 = 78.94 %...

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Explanation: