Answer:
[tex]\boxed{\text{40 mol Al}}[/tex]
Explanation:
Al₂O₃ ⟶ 2Al + 3O₂
n/mol: 20
[tex]\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}[/tex]
