Question

4.8 g O2 is mixed with 0.15 moles of iron. Given that iron (II) oxide will react with any available oxygen to
form iron (III) oxide completely, the mass of iron (III) oxide formed would be :
I will mark your answer the brainliest if you answer it properly.

Answer 1

Answer:

First Oxygen (O2) will react with Iron (Fe) to form Iron (II) oxide (FeO). When all the Iron (Fe) is converted to Iron (II) oxide (FeO), then FeO will react with the remaining (or available) Oxygen (O2) to produce Iron (III) oxide (Fe2O3).

The reaction between Oxygen (O2) and Iron (Fe) to form Iron (II) oxide (FeO) is:

Fe + O2 ----> 2FeO

(1 mole of Iron (Fe) reacts with 1 mole of Oxygen (O2) to form 2 moles of FeO)

Now, Number of moles of Fe = 0.15 moles

Number of moles of 4.8 g of O2 = (Given mass)/(Molecular mass) = 4.8/32 = 0.15 moles.

So, 0.15 moles of Iron (Fe) reacts with 0.15 moles of Oxygen (O2) to form 0.3 moles of FeO.

Therefore, moles of Oxygen (O2) left (or available) to react with Iron (II) oxide (FeO) to produce Iron (III) oxide (Fe2O3) = 0

Thus, Moles of Iron (III) oxide (Fe2O3) produced = 0

Hence, Mass of Iron (III) oxide (Fe2O3) is zero. (i.e. No Iron (III) oxide (Fe2O3) is formed