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#?

1 Answer
Apr 12, 2018

Answer:

#5 1/3 \ "mol of" \ AlCl_3#

Explanation:

This is an example of a limiting reactant problem.

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

#2Al(s)+3Cl_2(g)->2AlCl_3(s)#

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 1/3 \ "mol"# of aluminium chloride is formed.