`4Fe+3O_2 rarr 2Fe_2O_3`

To get 2 moles of `Fe_2O_3` we need 4 moles of Fe and 3 moles of `O_2` .

In the given mixture we have 4 moles of Fe and 2.5 moles of `O_2` .

Since we need 3 moles of `O_2` and 4 moles of Fe to get 2 moles of `Fe_2O_3` , the above reaction will not completely takes place in the mixture to form 2 moles of `Fe_2O_3` .

So what will happen is the reaction will happen until all the `O_2` is reacted.

So finally 2.5 moles of `O_2` will react.

`O_2:Fe_2O_3 = 3:2`

Amount of `Fe_2O_3` formed `= 2/3xx2.5 = 1.667`

So 1.667 moles of `Fe_2O_3` will form in the mixture.