Moles of Al₂O₃ formed : 1.42
Further explanation
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
