4Al + 3O2 = 2Al2O3. If 41.5 g aluminum and 69.0 g oxygen undergo a reaction that has a 67.0% yield, What mass of aluminum oxide forms. Please help by showing steps to find answers. Thanks

This is a limiting reagent problem. You know that because amounts for BOTH reactants are given. I solve these problems by working TWO stoichiometry problems. Here is a worked example for that. Print this and save it. It will solve 99.9% of the stoichiometry problems you will have.

http://www.jiskha.com/science/chemistry/stoichiometry.html
Use this method to solve for moles Al2O3 formed with 41.5 g Al. Work a second problem for Al3O3 formed with 69.0 g oxygen. Two answers but in limiting reagent problems the smaller value is ALWAYS the correct one and the reagent producing that value is the limiting reagent. The moles you find will be the theoretical yield (;i.e., 100%). Convert to grams and multiply by 0.67 since it is only 67% efficient.

To find the mass of aluminum oxide formed, you can follow these steps:

Step 1: Calculate the limiting reactant:
To determine the limiting reactant, we need to compare the moles of both reactants. First, calculate the number of moles for each substance:

Number of moles of aluminum (Al):
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 41.5 g / 26.98 g/mol = 1.537 mol

Number of moles of oxygen (O2):
Molar mass of O2 = 32.00 g/mol (Oxygen is diatomic)
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 69.0 g / 32.00 g/mol = 2.156 mol

Next, consider the balanced equation:

4Al + 3O2 → 2Al2O3

According to the stoichiometry, the ratio between Al and O2 is 4:3. This means that 4 moles of Al will react with 3 moles of O2.

Compare the calculated mole ratio between Al and O2:

Moles of Al / Coefficient of Al = Moles of O2 / Coefficient of O2
1.537 mol / 4 = 2.156 mol / 3

Simplifying the equation:
0.384 mol = 0.718 mol

Since you need more oxygen to react with aluminum (O2 moles > Al moles), aluminum is the limiting reactant.

Step 2: Calculate the theoretical yield:
Theoretical yield is the maximum amount of product that can be formed based on stoichiometric calculations.

From the balanced equation, we know that 4 mol Al reacts to form 2 mol of Al2O3.

Moles of Al2O3 = (Moles of Al / Coefficient of Al) * (Coefficient of Al2O3 / Coefficient of Al)

Moles of Al2O3 = (1.537 mol / 4) * (2 / 4)
Moles of Al2O3 = 0.384 mol

Step 3: Calculate the actual yield:
Actual yield is the yield obtained in the experiment, which is given as 67.0% of the theoretical yield.

Actual yield = Theoretical yield * Percentage yield / 100
Actual yield = 0.384 mol * 67.0 / 100
Actual yield = 0.257 mol

Step 4: Calculate the mass of Al2O3 formed:
To calculate the mass of Al2O3 in grams, multiply the number of moles of Al2O3 by its molar mass.

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

Mass of Al2O3 = Moles of Al2O3 x Molar mass of Al2O3
Mass of Al2O3 = 0.257 mol x 101.96 g/mol
Mass of Al2O3 = 26.22 g

Therefore, the mass of aluminum oxide formed is 26.22 grams.