How many moles of Al2O3 are formed from the reaction of 6.38 mol O2 and 9.15 mol Al? With Explanation please !

Moles of Al₂O₃ formed : 1.42

Given

6.33 mol O₂

9.15 mol Al

Required

moles of Al₂O₃

Solution

Reaction

4Al + 3O₂ ⇒ 2Al₂O₃

Find limiting reactant (mol : coefficient) :

Al = 9.15 : 4 = 2.29

O = 6.38 : 3 = 2.13

O₂ as a limiting reactant(smaller ratio)

Mol Al₂O₃ based on mol O₂ as a limiting reactant

From equation, mol ratio Al : Al₂O₃ = 4 : 2, so mol Al₂O₃ :

= 2/3 x mol O₂

= 2/3 x 2.13

= 1.42

answerto predict;