When 21.6g of aluminum reacts with oxygen, 20.8g of aluminum oxide is produced. what is the percentage yield of aluminum oxide?

To calculate the percentage yield of aluminum oxide, we need to compare the actual yield (20.8g) to the theoretical yield.

The balanced chemical equation for the reaction between aluminum and oxygen is:

4Al + 3O2 -> 2Al2O3

From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide.

First, we need to calculate the number of moles of aluminum oxide produced:

Molar mass of Al2O3 = (2 x atomic mass of Al) + (3 x atomic mass of O)
= (2 x 26.98 g/mol) + (3 x 16.00 g/mol)
= 101.96 g/mol

Number of moles of Al2O3 = mass / molar mass
= 20.8 g / 101.96 g/mol
= 0.204 moles

Next, we need to calculate the theoretical yield of aluminum oxide:

From the balanced equation, we can see that 4 moles of aluminum react to produce 2 moles of aluminum oxide. Therefore, the molar ratio is 2:4 or 1:2.

Number of moles of Al = 21.6 g / 26.98 g/mol
= 0.800 moles

Theoretical yield of Al2O3 = (0.800 moles Al) x (2 moles Al2O3 / 4 moles Al) x (101.96 g/mol)
= 0.400 moles x 101.96 g/mol
= 40.78 g

Finally, we can calculate the percentage yield:

Percentage yield = (actual yield / theoretical yield) x 100%
= (20.8 g / 40.78 g) x 100%
= 50.9%

Therefore, the percentage yield of aluminum oxide is approximately 50.9%.