Question

If 1210.5 g of copper (II) chloride ( $CuC{l_2}$ ) reacts with 510.0 g aluminium metal (Al) to produce 525.0 g copper metal (Cu), what is the limiting reactant?

Answer 1

The limiting reagent is the compound that is fully consumed in a chemical reaction. It is also known as Limiting reactant or limiting reagent. The product formed depends on the limiting reagent, since without it the reaction cannot continue.

Complete Step By Step Answer:
First we’ll find the balanced chemical equation for the given reaction. Th equation would be:
$2A{l_{(s)}} + 3CuC{l_{2(aq)}} \to 2AlC{l_{3(aq)}} + 3C{u_{(s)}}$
We will use the concept of stereochemistry to solve this problem. Here we can see that 2 moles of Al react with 3 moles of copper (II) chloride to form 2 moles of aluminium chloride and 3 moles of copper (solid).
Regardless of the amount of substance given to us in grams, we’ll always remember these mole ratios. First let us determine the no. of moles of each substance given to us. The no. of moles of any substance can be found out using the equation: $Moles = \dfrac{{mass(g)}}{{molar{\text{ }}mass(g/mol)}}$
The no. of moles of $CuC{l_2}$ given to us is (Molar mass = 134.45g/mol) $= \dfrac{{1210.5}}{{134.45}} = 9.003mol$
No. of moles of Aluminium metal given to us (molar mass = 26.982 g/mol) $= \dfrac{{510.0}}{{26.982}} = 18.90mol$
No. of moles of Copper metal produced (molar mass = 36.546 g/mol) $= \dfrac{{525.0}}{{36.546}} = 8.262mol$
We can see that the molar ratio of copper (II) chloride and copper is 3:3 = 1:1 i.e., no. of moles of $CuC{l_2}$ is equal to the no. of moles of copper, which is not seen from the above calculated no. of moles. This means that not all of the $CuC{l_2}$ has reacted with Al and has converted into copper. Hence, only 8.262 moles of copper (II) chloride has reacted with Al to form equivalent moles of copper (solid).
Now, we know that 8.262 moles of Copper (II) chloride has reacted with aluminium in the mole ratio of 2:3, let us find the moles of aluminium required for reacting $CuC{l_2}$
$3mol{\text{ }}CuC{l_2} = 2mol{\text{ }}Alu\min ium$
$18.90mol{\text{ }}Al = \dfrac{{18.90 \times 3}}{2} = 28.35mol$ of $CuC{l_2}$
As we can see that we are nowhere near to the no. of the moles actually required to react with 18.90 moles of Al. $CuC{l_2}$ is in the Limiting amount. Hence $CuC{l_2}$ is the Limiting Reagent.
Let us find the no. of moles of Al required to react with 8.262 mol of $CuC{l_2}$
$3mol{\text{ }}CuC{l_2} = 2mol{\text{ }}Alu\min ium$
$8.262mol{\text{ }}CuC{l_2} = \dfrac{{8.262 \times 2}}{3} = 5.508mol$ of Aluminium
The remaining moles of Aluminium reactant (apart from the 5.508 moles) are in excess.

Note:
The two methods followed to find the limiting agent are:
i) By comparing the amount of reactants. This is useful in the reaction where two reactants are present. In this the balanced equation is used to determine the amount of another reactant, by considering the first. If the second is more than the first, the first is the limiting reagent.
ii) By comparing the amount of product formed. The amount of product formed from the reactant present, is determined by the chemical equation. The reactant that forms the smallest amount of product is the limiting reagent.