16. Aluminum chloride, AlCl 3, is made by treating scrap aluminum with chlorine. 2 Al(s) + 3 Cl2(g) → 2 AlCl 3(s) If you begin with 2.70 g of Al and 4.05 g of Cl2,
(a) Which reactant is limiting?
2 moles of Al to 3 moles Cl₂
use the molar masses to find the number of moles
Cl2
To determine the limiting reactant, we need to compare the amount of product that can be formed from each reactant.
We'll use the following steps:
Step 1: Convert the given mass of Al to moles.
The molar mass of Al is 26.98 g/mol.
Number of moles of Al = Given mass (g) / Molar mass (g/mol)
Number of moles of Al = 2.70 g / 26.98 g/mol = 0.1001 mol
Step 2: Convert the given mass of Cl2 to moles.
The molar mass of Cl2 is 70.90 g/mol.
Number of moles of Cl2 = Given mass (g) / Molar mass (g/mol)
Number of moles of Cl2 = 4.05 g / 70.90 g/mol = 0.0572 mol
Step 3: Use the balanced equation to determine the mole ratio of Al to Cl2.
From the balanced equation: 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s), the mole ratio of Al to Cl2 is 2:3.
Step 4: Determine the limiting reactant.
Compare the moles of each reactant with the mole ratio.
For Al:
Moles of Al = 0.1001 mol
For Cl2:
Moles of Cl2 = 0.0572 mol
Since the mole ratio of Al to Cl2 is 2:3, we can calculate the number of moles of AlCl3 that can be formed from each reactant.
Moles of AlCl3 from Al = (0.1001 mol Al) × (2 mol AlCl3 / 2 mol Al) = 0.1001 mol AlCl3
Moles of AlCl3 from Cl2 = (0.0572 mol Cl2) × (2 mol AlCl3 / 3 mol Cl2) ≈ 0.0381 mol AlCl3
From the calculations, we can see that the moles of AlCl3 obtained from Al is higher than the moles obtained from Cl2. Therefore, the limiting reactant is Cl2.
To determine which reactant is limiting, we need to compare the amount of product that can be produced from each reactant.
Let's start by calculating the moles of aluminum, Cl2, and AlCl3 using their respective molar masses.
The molar mass of Al (aluminum) is 26.98 g/mol.
The molar mass of Cl2 (chlorine) is 70.90 g/mol.
The molar mass of AlCl3 (aluminum chloride) is 133.34 g/mol.
First, calculate the moles of aluminum:
moles of Al = mass of Al / molar mass of Al
moles of Al = 2.70 g / 26.98 g/mol
moles of Al ≈ 0.100 mol
Next, calculate the moles of Cl2:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 4.05 g / 70.90 g/mol
moles of Cl2 ≈ 0.057 mol
Now, let's compare the moles of each reactant to the stoichiometry of the balanced equation.
From the balanced equation: 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
We see that the molar ratio between Al and Cl2 is 2:3.
To determine how many moles of AlCl3 can be formed, we can use the mole ratio of Al to AlCl3:
moles of AlCl3 = (moles of Al) × (moles of AlCl3 / moles of Al)
moles of AlCl3 = 0.100 mol × (2 mol AlCl3 / 2 mol Al)
moles of AlCl3 = 0.100 mol
Similarly, we can determine how many moles of AlCl3 can be formed from Cl2:
moles of AlCl3 = (moles of Cl2) × (moles of AlCl3 / moles of Cl2)
moles of AlCl3 = 0.057 mol × (2 mol AlCl3 / 3 mol Cl2)
moles of AlCl3 ≈ 0.038 mol
Comparing the moles of AlCl3 that can be formed from each reactant, we see that the limiting reactant is the one that produces less moles of AlCl3.
In this case, the moles of AlCl3 that can be formed from Al is 0.100 mol, while the moles of AlCl3 that can be formed from Cl2 is only 0.038 mol.
Therefore, the limiting reactant is Cl2.