Question

A certain oxide of iron contains 2.5 grams of oxygen
for every 7.0 grams of iron. If it is regarded as a
mixture of FeO and Fe,0, in the weight ratio a: b,
what is a:b? (atomic weight of iron = 56)
fol
. 10
hi
90​

Answer 1

x : y = 0.9 : 1.0

Explanation:

Let us take p as moles of Fe +2, q as moles of Fe +3.

The oxide has p, q, 1 mole of O.

For every 2.5 gram of O, we have 7 gm Fe.

For every 16 gram of O, we have (7/2.5) x 16 gm Fe.

So for every mole of O, we have (7 x 16) / (2.5 x 56) moles of Fe, or 0.8 moles of Fe.

In order for the molecule to be electrically neutral:

   2p +3q = 2

   2p + 3(0.8 - p) = 2

   Substituting p = 0.4, we get q = 0.4.

So the oxide has 0.4 moles of Fe +2, 0.4 moles of Fe +3 and 1 mole of O. We can divide the 1 mole of oxygen atom as 0.4 mole and 0.6 mole and say there are 0.4 moles of FeO and 0.2 moles of Fe2O3.

Mass of 1 mole of FeO is 72 grams. Hence, mass of 0.4 moles of FeO has a mass of 28.8 ----- (x)

Mass of 1 mole of Fe2O3 is 160 grams. Hence, mass of 0.2 moles of Fe2O3 has a mass of 32 ----- (y)

So  ratio of x and y = x : y = 28.8 : 32 = 0.9 : 1.0

Answer 2

Explanation:

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