# If 1210.5 g of copper (II) chloride (CuCl2) reacts with 510.0 g aluminum metal (Al) to produce 525.0 g copper metal (Cu), what is the limiting reactant?

Apr 22, 2015

The copper (II) chloride will be your limiting reagent.

$\textcolor{red}{2} A {l}_{\left(s\right)} + \textcolor{b l u e}{3} C u C {l}_{2 \left(a q\right)} \to 2 A l C {l}_{3 \left(a q\right)} + \textcolor{b r o w n}{3} C {u}_{\left(s\right)}$

The most important tool you'll have at your disposal to solve this problem (or any stoichiometry problem, for that matter) is the mole ratio. Notice that you need $\textcolor{red}{2}$ moles of aluminium to react with $\textcolor{b l u e}{3}$ moles of copper (II) chloride to produce $\textcolor{b r o w n}{3}$ moles of copper.

Regardless of what the exact number of moles of each you have, they always must respect these ratios.

So, the first thing to do is determine how many moles of each species you're dealing with. To do this, use the relationship that exists between mass and number of moles

1210.5cancel("g") * ("1 mole "CuCl_2)/(134.45cancel("g")) = "9.003 moles " $C u C {l}_{2}$

510.0cancel("g") * "1 mole Al"/(26.982cancel("g")) = "18.90 moles Al"

525.0cancel("g") * "1 mole Cu"/(63.546cancel("g")) = "8.262 moles Cu"

The first thing that stands out is the fact that the number of moles of copper (II) chloride is not equal to the number of moles of copper, despite the fact that they have a $1 : 1$ ($\textcolor{b l u e}{3} : \textcolor{b r o w n}{3}$) mole ratio.

This means that not all the copper (II) chloride reacted with the aluminium metal. More specifically, only 8.262 moles of $C u C {l}_{2}$ reacted.

So, you know that only 8.626 moles of $C u C {l}_{2}$ have reacted. Now it's time to compared this number with the number of moles of aluminium present.

According to the $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio that exists between the two, you would need

18.90cancel("moles Al") * (color(blue)(3)" moles "CuCl_2)/(color(red)(2)cancel("moles Al")) = "28.35 moles " $C u C {l}_{2}$

Since you don't even come close to having this many moles of copper (II) chloride present, $C u C {l}_{2}$ will act as a limiting reagent. As a result, the number of moles of aluminium that react will be

8.262cancel("moles "CuCl_2) * (color(red)(2)" moles Al")/(color(blue)(3)cancel("moles "CuCl_2)) = "5.508 moles Al"

The rest of the aluminium will be in excess.

SIDE NOTE When I said that not all the copper (II) chloride reacted, I assumed that your reaction had a 100% percent yield.

I suspect that this wasn't the case, and a follow-up question would be to calculate either the theoretical yield of copper or the percent yield of the reaction.