Theoretical yield of Al₂O₃: 1.50 mol.
[tex]2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3[/tex];
[tex]4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}[/tex].
How many moles of aluminum oxide formula units will be produced if aluminum is the limiting reactant?
Aluminum reacts to aluminum oxide at a two-to-one ratio.
[tex]3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}[/tex].
As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced if oxygen is the limiting reactant?
Oxygen reacts to produce aluminum oxide at a three-to-two ratio.
[tex]2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}[/tex]
As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.
How many moles of aluminum oxide formula units will be produced?
Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.
