Aluminum and oxygen react to form aluminum oxide.

4 Al + 3 O2 2 Al2O3



a. What is the limiting reactant if 36.4 g Al are reacted with 36.4 g of O2?

O2



b. What is the theoretical yield of Al2O3, in grams?

68.8 g



c. If 22.2 g of Al2O3 are obtained, what is the percent yield?

32.3

I think the limiting reagent is Al.

Your 68.8g, however, is correct for theoretical yield of Al2O3 which means you must have used 36.4g Al to solve for g Al2O3.
The limiting reagent is the substance that is completely used.
%yield is correct.

To determine the limiting reactant in a chemical reaction, you need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.

a. To find the limiting reactant, you can use the concept of molar ratios. Convert the given masses of Al and O2 to moles using their respective molar masses. The molar mass of Al is 26.98 g/mol, and the molar mass of O2 is 32.00 g/mol.

For Al, moles = (mass of Al) / (molar mass of Al) = 36.4 g / 26.98 g/mol = 1.35 mol

For O2, moles = (mass of O2) / (molar mass of O2) = 36.4 g / 32.00 g/mol = 1.14 mol

Next, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. From the balanced equation, we can see that the stoichiometric ratio is 4:3 for Al to O2.

The ratio of moles of Al to moles of O2 = 1.35 mol Al / 1.14 mol O2 = 1.18.

Since the ratio is greater than 1, it means that there is excess O2 left after the reaction when 36.4 g of Al reacts with 36.4 g of O2. Therefore, the limiting reactant is O2.

b. To calculate the theoretical yield of Al2O3, use the stoichiometric ratio from the balanced equation. From the balanced equation, we can see that 4 moles of Al produce 2 moles of Al2O3.

The moles of Al2O3 produced can be calculated as follows:

moles of Al2O3 = (moles of Al) x (moles of Al2O3 / moles of Al)
= 1.35 mol Al x (2 mol Al2O3 / 4 mol Al)
= 0.675 mol Al2O3

Now, convert moles of Al2O3 to grams using the molar mass of Al2O3. The molar mass of Al2O3 is 101.96 g/mol.

mass of Al2O3 = (moles of Al2O3) x (molar mass of Al2O3)
= 0.675 mol Al2O3 x 101.96 g/mol
= 68.8 g

Therefore, the theoretical yield of Al2O3 is 68.8 grams.

c. To calculate the percent yield, divide the actual yield (22.2 g) by the theoretical yield (68.8 g) and multiply by 100.

percent yield = (actual yield / theoretical yield) x 100
= (22.2 g / 68.8 g) x 100
= 32.3%

Therefore, the percent yield of Al2O3 is 32.3%.