# How do you determine the maximum number of moles of product that can be produced from 7.0 mol Al and 8.0 mol Cl_2 according to the equation 2Al + 3Cl_2 -> 2AlCl_3?

Apr 12, 2018

$5 \frac{1}{3} \setminus \text{mol of} \setminus A l C {l}_{3}$

#### Explanation:

This is an example of a limiting reactant problem.

The first step is to write down the balanced chemical equation. That is,

$2 A l \left(s\right) + 3 C {l}_{2} \left(g\right) \to 2 A l C {l}_{3} \left(s\right)$

We got seven moles of aluminium and eight moles of chlorine. So, we find the limiting reactant.

7color(red)cancelcolor(black)("mol of" \ Al)*(2 \ "mol of" \ AlCl_3)/(2color(red)cancelcolor(black)("mol of" \ Al))=7 \ "mol of" \ AlCl_3

8color(red)cancelcolor(black)("mol of" \ Cl_2)*(2 \ "mol of" \ AlCl_3)/(3color(red)cancelcolor(black)("mol of" \ Cl_2))=5 1/3 \ "mol of" \ AlCl_3

So, this shows that chlorine gas is the limiting reactant here, and only the maximum amount of aluminium chloride will form is dependent on the amount of chlorine gas.

Since there are eight moles of chlorine gas, then $5 \frac{1}{3} \setminus \text{mol}$ of aluminium chloride is formed.