The theoretical yield : = 28.34 g Al₂O₃
Further explanation
Given
15 g of Aluminum react
Required
The theoretical yield
Solution
Reaction
4Al(s) + 3O₂ (g)⇒ 2Al₂O₃(s)
mol of Aluminum :
mol = mass : Ar Al
mol = 15 g : 26.98 g/mol
mol = 0.556
From the equation, mol ratio of Al : Al₂O₃ = 4 : 2, so mol Al₂O₃ :
= 2/4 x mol Al
= 2/4 x 0.556
= 0.278
Mass Al₂O₃ :
= mol x MW Al₂O₃
= 0.278 x 101.96
= 28.34 g
