Answer:
34.2%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Cr₂O₃ + 2Al —> Al₂O₃ + 2Cr
Next, we shall determine the mass of Cr₂O₃ that reacted and the mass of Al₂O₃ produced from the balanced equation. This can be obtained as follow:
Molar mass of Cr₂O₃ = (52×2) + (3×16)
= 104 + 48
= 152 g/mol
Mass of Cr₂O₃ from the balanced equation = 152 × 1 = 152 g
Molar mass of Al₂O₃ = (27×2) + (16×3)
= 54 + 48
= 102 g/mol
Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g
SUMMARY:
From the balanced equation above,
152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.
Next, we shall determine the theoretical yield of Al₂O₃. This can be obtained as follow:
From the balanced equation above,
152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.
Therefore, 23.3 g of Cr₂O₃ will react to produce = (23.3 × 102)/152 = 15.64 g of Al₂O₃.
Thus, the theoretical yield of Al₂O₃ is 15.64 g.
Finally, we shall determine the percentage yield of Al₂O₃. This can be obtained as follow:
Actual yield of Al₂O₃ = 5.35 g
Theoretical yield of Al₂O₃ = 15.64 g.
Percentage yield of Al₂O₃ =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 5.35 / 15.64 × 100
Percentage yield of Al₂O₃ = 34.2%
