Question

using the balance equation2 Al +3 Cl^2 arrow 2AlCl^3if you use 54 grams of aluminum and 140 grams of chlorine, which is the limiting reactant?

Answer 1

Answer:

Limiting reactant = Cl2.

Explanation:

First, let's review the concept of limiting reactant: the limiting reactant is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Our balanced chemical equation is:

$2Al+3Cl_2\rightarrow2AlCl_3.$

Let's convert 54 g of aluminum (Al) to moles using its molar mass which is 27 g/mol:

$54\text{ g Al}\cdot\frac{1\text{ mol Al}}{27\text{ g Al}}=2\text{ moles Al.}$

You can see in the chemical equation, that 2 moles of Al produce 2 moles of AlCl3. This is telling us that the molar ratio between Al and AlCl3 is 1:1, so we're producing 2 moles of Al:

$2\text{ moles Al}\cdot\frac{2\text{ moles AlCl}_3}{2\text{ moles Al}}=2\text{ moles AlCl}_3.$

Now, let's see how many moles of AlCl3 we can produce from 140 g of chlorine (Cl2). The molar mass of Cl2 is 70.8 g/mol:

$140\text{ g Cl}_2\cdot\frac{1\text{ mol Cl}_2}{70.8\text{ g Cl}_2}=1.97\text{ moles Cl}_2.$

The next step is to see how many moles of AlCl3 are being produced. You can see in the chemical equation that 3 moles of Cl2 produce 2 moles of AlCl3:

$1.97\text{ moles Cl}_2\cdot\frac{2\text{ moles AlCl}_3}{3\text{ moles Cl}_2}=1.31\text{ moles AlCl}_3.$

In this case, our limiting reactant is Cl2, because it limits the amount of product that can be produced because 1.31 moles of AlCl3 is less than 2 moles of AlCl3.