Aluminum and oxygen react to form aluminum oxide.

4 Al + 3 O2 2 Al2O3

a. What is the limiting reactant if 30.7 g Al are reacted with 38.3 g of O2?

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

c. If 35.3 g of Al2O3 are obtained, what is the percent yield?

To determine the limiting reactant, theoretical yield, and percent yield for a chemical reaction, we need to follow a few steps. Let's analyze each part of the question:

a. Limiting Reactant:
1. Convert the given masses of Aluminum (Al) and oxygen (O2) to moles by using their molar masses. The molar mass of Aluminum is 26.98 g/mol, and the molar mass of oxygen is 32.00 g/mol.
- Moles of Al = 30.7 g Al / 26.98 g/mol = 1.14 mol Al
- Moles of O2 = 38.3 g O2 / 32.00 g/mol = 1.1975 mol O2

2. Divide the moles of each reactant by its stoichiometric coefficient. In this balanced equation, the coefficients of Al and O2 are 4 and 3, respectively.
- Moles of Al / coefficient of Al = 1.14 mol Al / 4 = 0.285 mol Al
- Moles of O2 / coefficient of O2 = 1.1975 mol O2 / 3 = 0.39925 mol O2

3. The limiting reactant is the one with the smaller value obtained in step 2. In this case, Aluminum (Al) has the smaller value (0.285 mol), so Aluminum is the limiting reactant.

b. Theoretical Yield:
1. Determine the number of moles of Aluminum Oxide (Al2O3) produced from the limiting reactant, Aluminum (Al). From the balanced equation, we know that 4 moles of Al react to form 2 moles of Al2O3.
- Moles of Al2O3 = Moles of Al * (Ratio of Al2O3 / Al) = 0.285 mol Al * (2 mol Al2O3 / 4 mol Al) = 0.1425 mol Al2O3

2. Calculate the theoretical yield in grams by multiplying the moles of Al2O3 by its molar mass. The molar mass of Al2O3 is 101.96 g/mol.
- Theoretical yield of Al2O3 = 0.1425 mol Al2O3 * 101.96 g/mol = 14.53 g Al2O3

c. Percent Yield:
1. Convert the given mass of Al2O3 obtained to moles using its molar mass.
- Moles of Al2O3 obtained = 35.3 g Al2O3 / 101.96 g/mol = 0.3461 mol Al2O3

2. Calculate the percent yield by dividing the actual yield (moles of Al2O3 obtained) by the theoretical yield, and then multiplying by 100%.
- Percent yield = (Moles of Al2O3 obtained / Moles of Al2O3 expected) * 100%
- Percent yield = (0.3461 mol Al2O3 / 0.1425 mol Al2O3) * 100% ≈ 242.7%

So, the answers to the questions are:

a. The limiting reactant is Aluminum (Al).
b. The theoretical yield of Al2O3 is 14.53 grams.
c. The percent yield is approximately 242.7%.