Question

Calculate the volume of O2 required for the production of 0.3 moles of Fe2O3​

Answer 1

Answer:

10.08 liters of O2 are required for producing 0.3 moles of Fe2O3

Explanation:

4 Fe + 3 O2 ---> 2 Fe2O3

3 moles of O2 = 2 Moles Fe2O3

? Moles of O2 = 0.3 moles of Fe2O3

Moles of O2 = (0.3 × 3)/2 = 0.9/2 = 0.45 moles of O2

1 mole = 22.4l of O2 at STP

0.45 moles = ? liters of O2

0.45 × 22.4 = 10.08 liters of O2

Answer 2

Therefore, 10.09 Litre of Oxygen is required for the production of 0.3 moles of Fe₂O₃.

The volume of O₂ required for the production of 0.3 moles of Fe₂O₃ are discussed below -

Given:  Number of moles of Fe₂O₃ = 0.3

To find The volume of oxygen required for the production of 0.3 Fe₂O₃

Solution:

The balanced chemical reaction which takes place for the formation of Fe₂O₃ are -

4Fe + 3O₂ ---> 2Fe₂O₃

So from the law of conservation of mass, three moles of oxygen is required to produce two moles of Fe₂O₃.

So, 0.3 moles of Fe₂O₃ will require 3/2×0.3 moles of oxygen.

Therefore, 0.45 moles of oxygen is required to produce 0.3 moles of Fe₂O₃

Now as we know, 1 mole of any element occupies 22.4 Litre of volume.

Therefore, 0.45 moles of Oxygen occupies 22.4×0.45 L of volume.

Therefore, 10.09 L of Oxygen is required for the production of 0.3 moles of Fe₂O₃.