# How many chloride ions are there in 4.50 mol of aluminum chloride?

May 2, 2018

$\text{13.5mol} \left(C {l}^{+}\right)$

#### Explanation:

Aluminum Chloride is $A l C {l}_{3}$.

For every moles of $A l C {l}_{3}$, you have 3 moles of $C {l}^{+}$ ions.

Therefore:

$\text{4.50mol" (AlCl_3) * ("3mol" (Cl))/("1mol" (AlCl_3)) = "13.5mol} \left(C l\right)$