How many grams of oxygen are needed to form 122 g of aluminum oxide?

Mass of O₂ needed : = 57.408 g

Given

122 g of Aluminum oxide

Required

Mass of Oxygen

Solution

Reaction

4Al + 3O₂ → 2Al₂O₃

Mol Al₂O₃(MW=102 g/mol) :

= mass : MW

= 122 g : 102 g/mol

= 1.196

From the equation, mol ratio of O₂ : Al₂O₃ = 3 : 2, so mol O₂ :

= 3/2 x mol Al₂O₃

= 3/2 x 1.196

= 1.794

Mass O₂ (MW = 32 g/mol) :

= mol x MW

= 1.794 x 32

= 57.408 g