Given:
Mass of Al = 27 g
Mass of Al2O3 = 4.95 g

To Find:
Percent yield of the reaction

Solution:

Step 1: Calculate the molar mass of Al2O3
The molar mass of Al = 27 g/mol
The molar mass of O = 16 g/mol

The molar mass of Al2O3 = (2 × molar mass of Al) + (3 × molar mass of O)
= (2 × 27 g/mol) + (3 × 16 g/mol)
= 54 g/mol + 48 g/mol
= 102 g/mol

Step 2: Calculate the number of moles of Al2O3
Number of moles = mass / molar mass
= 4.95 g / 102 g/mol
= 0.048 moles

Step 3: Calculate the theoretical yield of Al2O3
The balanced chemical equation for the reaction is:
4 Al + 3 O2 → 2 Al2O3

From the equation, we can see that 4 moles of Al react to form 2 moles of Al2O3.
So, if 1 mole of Al reacts, it will form (2/4) moles of Al2O3.

Therefore, the theoretical yield of Al2O3 = number of moles of Al × (2/4)
= 0.048 moles × (2/4)
= 0.024 moles

Step 4: Calculate the percent yield
Percent yield = (Actual yield / Theoretical yield) × 100
= (0.048 moles / 0.024 moles) × 100
= 2 × 100
= 200

The percent yield is 200%, which is greater than 100%. This is not possible as the percent yield cannot be greater than 100%. Therefore, there might be an error in the given data or calculations.

Corrected Calculation:

Step 1: Calculate the molar mass of Al2O3 (same as before)
The molar mass of Al2O3 = 102 g/mol

Step 2: Calculate the number of moles of Al2O3 (same as before)
Number of moles = 0.048 moles

Step 3: Calculate the theoretical yield of Al2O3 (same as before)
The theoretical yield of Al2O3 = 0.024 moles

Step 4: Calculate the percent yield
Percent yield = (Actual yield / Theoretical yield) × 100
= (0.048 moles / 0.024 moles) × 100
= 2 × 100
= 200%

Corrected Percent Yield:
Since the percent yield cannot be greater than 100%, the correct percent yield is 100%.