To calculate the theoretical yield of alum from aluminum foil, we can start with the balanced chemical equation for the formation of alum, which can be represented as:
2 K⁺ + Al³⁺ + 2 SO₄²⁻ + 12 H₂O → K₂SO₄ • Al₂(SO₄)₃ • 24 H₂O (Alum)
From the equation, we can see that 1 mole of Aluminum (Al) yields 1 mole of Alum. First, we need to find out how many moles of Al are present in 1.0000 g of aluminum foil.
The molar mass of Aluminum (Al) is approximately 26.98 g/mol. Thus, the number of moles of Al in 1.0000 g is calculated as follows:
Number of moles = Mass (g) / Molar mass (g/mol) = 1.0000 g / 26.98 g/mol ≈ 0.0370 moles
Since the reaction indicates that 1 mole of Al produces 1 mole of Alum, we can say that 0.0370 moles of Aluminum will produce 0.0370 moles of Alum.
Next, we need to calculate the theoretical yield of alum. The molar mass of alum (K₂SO₄ • Al₂(SO₄)₃ • 24 H₂O) is approximately 474.39 g/mol. Therefore, the theoretical yield can be determined by:
Theoretical yield (g) = Number of moles × Molar mass = 0.0370 moles × 474.39 g/mol ≈ 17.55 g
Now, let’s compute the percent yield based on the actual yield recovered, which is given as 16.2105 g. The percent yield can be calculated using the formula:
Percent yield = (Actual yield / Theoretical yield) × 100%
Substituting the values we have:
Percent yield = (16.2105 g / 17.55 g) × 100% ≈ 92.35%
In conclusion, based on 1.0000 g of aluminum foil, the theoretical yield of alum is approximately 17.55 g, and the percent yield based on the actual recovery of 16.2105 g is approximately 92.35%.