Answer:
[tex]\rm 2.69\ mol\ Al_2O_3[/tex]
Explanation:
We are applying stoichiometry, or unit conversion, to the reaction:
[tex]\rm 4Al_{(s)} \ +\ 3O_2_{(g)} \to 2Al_2O_3_{(g)}[/tex]
We are given the initial amounts of reactants:
- [tex]\rm 7.52\ mol\ Al_{(s)}[/tex]
- [tex]\rm 4.03\ mol\ O_{2(g)}[/tex]
First, we need to find the limiting reactant. For the initial amount of aluminum, we would need:
[tex]\rm \dfrac{7.52\ mol\ Al}{1}\times \dfrac{3\ mol\ O_2}{4\ mol\ Al} = 5.64\ mol\ O_2\text{ needed}[/tex]
Since we have less oxygen than we need, O₂ is the limiting reactant. So, this is the one we will do stoichiometry with:
[tex]\rm \dfrac{4.03\ mol\ O_2}{1}\times\dfrac{2\ mol\ Al_2O_3}{3\ mol\ O_2} = \boxed{\rm 2.69\ mol\ Al_2O_3}[/tex]
