Answer:
[tex]\large \boxed{\text{2.52 mol Al}}[/tex]
Explanation:
2Al₂O₃ ⟶ 4Al +3O₂
n/mol: 1.26
The molar ratio is 4 mol Al:2 mol Al₂O₃.
[tex]\text{Moles of Al} = \text{1.26 mol Al$_{2}$O}_{3} \times \dfrac{\text{4 mol Al}}{\text{2 mol Al$_{2}$O}_{3}}= \textbf{2.52 mol Al}\\\\\text{The reaction produces $\large \boxed{\textbf{2.52 mol Al}}$}[/tex]
