Answer:
[tex]\large \boxed{\text{0.2 mol of Al$_{2}$O}_{3}}}[/tex]
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
2Al +3FeO ⟶ 3Fe + Al₂O₃
n/mol: 0.6
1 mol of Al₂O₃ is produced along with 3 mol of Fe
[tex]\text{Moles of Al$_{2}$O}_{3} = \text{0.6 mol Fe}\times \dfrac{\text{1 mol Al$_{2}$O}_{3}}{\text{3 mol Fe}}= \textbf{0.2 mol Al$_\mathbf{2}$O}_{\mathbf{3}}\\\\\text{The reaction will produce $\large \boxed{\textbf{0.2 mol of Al$_\mathbf{2}$O}_{\mathbf{3}}}$ }[/tex]
