Answer:
The balanced chemical equation is:
[tex]2Al+3FeCl_2\operatorname{\rightarrow}2AlCl_3+3Fe[/tex]And we will produce 1.2 moles of Fe from 0.80 moles of Al.
Explanation:
First, let's write the chemical equation:
[tex]Al+FeCl_2\rightarrow AlCl_3+Fe.[/tex]You can note that is unbalanced because in the reactants we have 2 moles of Cl and in the products, we have 3 moles of Cl. So if we put '3' moles beside FeCl2, we're going to obtain 6 moles of Cl and if we put '2' moles beside AlCl3, we're going to have 6 moles of Cl too:
[tex]Al+3FeCl_2\operatorname{\rightarrow}2AlCl_3+Fe[/tex]But you can note that it is still unbalanced for Al and Fe. We have 1 mol of Al in the reactants and 2 moles in the products. We have 3 moles of Fe in the reactants and 1 mol of Fe in the products. So putting '2' moles beside Al and '3' moles beside Fe, we obtain:
[tex]2Al+3FeCl_2\operatorname{\rightarrow}2AlCl_3+3Fe[/tex]And that would be the balanced equation.
Now, we want to find the number of moles of iron (Fe) produced by 0.80 moles of Al with an excess of iron (II) chloride. This means that we take into account the number of moles of the limiting reactant, in this case, 0.80 moles of Al (the limiting reactant).
You can note that in the balanced chemical equation, 2 moles of Al reacted produces 3 moles of iron (Fe), so let's state a rule of three to solve this problem, like this:
[tex]\begin{gathered} 2\text{ moles Al}\rightarrow3\text{ moles Fe} \\ 0.80\text{ moles Al}\rightarrow\text{? moles Fe.} \end{gathered}[/tex]And the calculation would look like this:
[tex]0.80\text{ moles Al}\cdot\frac{3\text{ moles Fe}}{2\text{ moles Al}}=1.2\text{ moles Fe.}[/tex]We will obtain 1.2 moles of iron (Fe) produced by 0.80 moles of Al reacted.