Question

In the oxidation of iron; how many grams of iron (III) oxide will be produced from 6.20 mol of Fe?

Answer 1

Answer:

496 g of Fe₂O₃.

Explanation:

The balanced equation for the reaction is given below:

4Fe + 3O₂ —> 2Fe₂O₃

From the balanced equation above,

4 moles of Fe reacted to produce 2 moles of Fe₂O₃.

Therefore, 6.20 moles of Fe will react to produce = (6.20 × 2)/4 = 3.1 moles of Fe₂O₃

Finally, we shall determine the mass of 3.1 moles of Fe₂O₃. This can be obtained as follow:

Mole of Fe₂O₃ = 3.1 moles

Molar mass of Fe₂O₃ = (56 × 2) + (3×16)

= 112 + 48

= 160 g/mol

Mass of Fe₂O₃ =?

Mass = mole × molar mass

Mass of Fe₂O₃ = 3.1 × 160

Mass of Fe₂O₃ = 496 g

Therefore, 496 g of Fe₂O₃ were produced from the reaction.