Let number of 02- = 100
number of fe2+ = x
number of fe3+ =  93 - x
to maintain electrical neutrality net positive charge is equal to net negative charge 
   2x = 3 (93 - x) = 2 * 100
   2x + 279 - 3x = 200
   x = 79
Fe²± = 79
Fe+2/Fe+2 + Fe+3 -----> 79/93= 0.849
Hope it helps!!!!!!

The fraction of iron present as Fe2+ in Fe0.90 can be calculated by considering the molar ratio between Fe2+ and Fe0.90.

To begin with, let's understand the concept of molar ratio. The molar ratio compares the number of moles of one substance to another in a chemical reaction. In this case, we want to determine the molar ratio between Fe2+ and Fe0.90.

The Molar Ratio:
The molar ratio between Fe2+ and Fe0.90 can be derived from the balanced chemical equation for the reaction:

2Fe0.90 + 2H+ → 2Fe2+ + H2

From the equation, we can see that for every 2 moles of Fe0.90, we obtain 2 moles of Fe2+. This indicates a 1:1 molar ratio between Fe0.90 and Fe2+.

Calculating the Fraction:
To calculate the fraction of iron present as Fe2+ in Fe0.90, we need to determine the moles of Fe2+ and Fe0.90.

Assuming we have 1 mole of Fe0.90, we can calculate the moles of Fe2+ using the molar ratio:

Moles of Fe2+ = Moles of Fe0.90

Since the molar ratio is 1:1, the moles of Fe2+ will also be 1.

Now we can calculate the fraction of iron present as Fe2+:

Fraction of Fe2+ = Moles of Fe2+ / Total moles of iron

Since we have 1 mole of Fe2+ and 1 mole of Fe0.90, the total moles of iron is 2.

Fraction of Fe2+ = 1 / 2 = 0.5

Conclusion:
The fraction of iron present as Fe2+ in Fe0.90 is approximately 0.5 or 50%. This means that half of the iron in Fe0.90 exists as Fe2+ ions.