Explanation:
The chemical equation of the reaction that occurs when [tex]AlF_{3}[/tex] reacts with [tex]O_{2}[/tex] is
[tex]4AlF_{3}+3O_{2}[/tex]→[tex]2Al_{2}O_{3}+6F_{2}[/tex]
[tex]4[/tex] moles of [tex]AlF_{3}[/tex] requires [tex]3[/tex] moles of [tex]O_{2}[/tex].
[tex]1[/tex] mole of [tex]AlF_{3}[/tex] requires [tex]\frac{3}{4}[/tex] moles of [tex]O_{2}[/tex].
Given that we have [tex]9[/tex] moles of [tex]AlF_{3}[/tex].
[tex]9[/tex] moles of [tex]AlF_{3}[/tex] requires [tex]\frac{3}{4}\times 9=6.75[/tex] moles of [tex]O_{2}[/tex].
But we have [tex]12[/tex] moles of [tex]O_{2}[/tex].
So,[tex]AlF_{3}[/tex] will be consumed first.
So,[tex]AlF_{3}[/tex] is the limiting reagent.
