Question

What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?

Answer 1

Answer:$AlF_{3}$

Explanation:

The chemical equation of the reaction that occurs when $AlF_{3}$ reacts with $O_{2}$ is

$4AlF_{3}+3O_{2}$→$2Al_{2}O_{3}+6F_{2}$

$4$ moles of $AlF_{3}$ requires $3$ moles of $O_{2}$.

$1$ mole of $AlF_{3}$ requires $\frac{3}{4}$ moles of $O_{2}$.

Given that we have $9$ moles of $AlF_{3}$.

$9$ moles of $AlF_{3}$ requires $\frac{3}{4}\times 9=6.75$ moles of $O_{2}$.

But we have $12$ moles of $O_{2}$.

So,$AlF_{3}$  will be consumed first.

So,$AlF_{3}$  is the limiting reagent.