Question

I really need help!!!

Calculate the number of moles of Al2 O3 that are produced when 0.60 mol of Fe is produced in the following reaction.
2Al(s) + 3FeO(s) arrow 3Fe(s) + Al 2 O3 (s)

a.0.20 molc.0.60 mol

b.0.40 mold.0.90 mol

Answer 1

The correct option is a

Step-by-step explanation:

In the reaction, 2 moles of Aluminium (Al) reacted with 3 moles of Iron (II) oxide (FeO) to produce Iron (Fe) and Aluminium oxide (Al₂O₃) in the ratio 3:1 respectively. From the above, it can be deduced that when 3 moles of iron is produced, 1 mole of aluminium oxide is produced, hence when 0.6 mole of iron is produced, X mole of aluminium oxide is produced, where X is the unknown.

3 ⇒ 1

0.6 ⇒ X

X =
`(0.6 * 1)/(3)`

X = 0.2 mole of Al₂O₃ will be produced when 0.60 mole of Fe is produced in the reaction (in the question)

Answer 2

We use the mole ratio of aluminum oxide Al2O3 and iron Fe from their coefficients in the balanced chemical equation
2Al(s) + 3FeO(s) = 3Fe(s) + Al2O3(s)
which is one mole of aluminum oxide is to react with three moles of iron.

Therefore, the number of moles of aluminum oxide is
moles of Al2O3 = 0.60 mol of Fe * 1 mol Al2O3 / 3 mol Fe
= 0.20 mol