Mass of O₂ needed : = 57.408 g
Further explanation
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