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?
1 Answer
The copper (II) chloride will be your limiting reagent.
Start with the balanced chemical equation for your reaction
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
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
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
This means that not all the copper (II) chloride reacted with the aluminium metal. More specifically, only 8.262 moles of
So, you know that only 8.626 moles of
According to the
Since you don't even come close to having this many moles of copper (II) chloride present,
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.