Anhydrous aluminum oxide can be reduced to aluminum according to thisa chemical equation: 2 Al2O3 (s) ---> 4 Al (l) + 3 O2 (g) Calculate the mass of aluminum oxide reacted if 82 g of Al was produced and the percent yield was found to be 89.3%
This works the same way as the Cu(NO3)2 problem.
First, what mass Al would we have had at 100% yield; i.e., the theoretical yield? Thats
%yield = [(actual yield/theoretical yield)]*100
89.3% = (82/theor)*100. Solve for theoretical yield and use that below.
mols Al = grams/molar mass.
Using the coefficients, convert mols Al to mols Al2O3.
Then convert mols Al2O3 t grams.
I ran through and estimated 193g Al2O3 needed to produce 82g Al at 89.3% yield.
Be sure to confirm that.
To find the mass of aluminum oxide reacted, we need to use the equation and the given information on the percent yield.
First, determine the molar mass of Al and Al2O3:
- The molar mass of Al is 26.98 g/mol.
- The molar mass of Al2O3 is (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol.
Next, we can set up a proportion to find the theoretical yield of Al2O3 based on the mass of Al produced:
(82 g Al) / (4 * 26.98 g Al/mol) = (x g Al2O3) / (101.96 g Al2O3/mol)
Simplifying the proportion:
82 g Al / (4 * 26.98 g Al/mol) = x g Al2O3 / (101.96 g Al2O3/mol)
Solving for x, the mass of Al2O3:
x = (82 g Al / (4 * 26.98 g Al/mol)) * (101.96 g Al2O3/mol)
x ≈ 189.36 g Al2O3
The calculated mass of aluminum oxide reacted is approximately 189.36 g.
To calculate the mass of aluminum oxide reacted, we can use the following steps:
Step 1: Calculate the theoretical yield of aluminum.
The molar ratio between aluminum oxide (Al2O3) and aluminum (Al) is 2:4, which means that for every 2 moles of aluminum oxide reacted, 4 moles of aluminum are produced.
Given that the molar mass of aluminum is 26.98 g/mol and the molar mass of aluminum oxide is (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol, we can calculate the theoretical yield of aluminum as follows:
(82 g Al) x (4 mol Al / 26.98 g Al) = 12.128 mol Al
Step 2: Calculate the actual yield of aluminum.
The percent yield is given as 89.3%, which represents the actual yield as a percentage of the theoretical yield. We can calculate the actual yield of aluminum as follows:
Actual yield = (Percent yield / 100) x Theoretical yield
Actual yield = (0.893) x (12.128 mol Al)
Actual yield = 10.848 mol Al
Step 3: Convert the actual yield of aluminum to mass.
To calculate the mass of aluminum oxide reacted, we need to determine the molecular weight of aluminum oxide and use it in the stoichiometric ratio.
The molecular weight of aluminum oxide (Al2O3) is (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol.
Mass of aluminum oxide reacted = (Actual yield of aluminum) x (Molar mass of aluminum oxide)
Mass of aluminum oxide reacted = (10.848 mol Al) x (101.96 g/mol Al2O3)
Mass of aluminum oxide reacted = 1,105.914336 g Al2O3
Therefore, the mass of aluminum oxide reacted is approximately 1,105.914 g.