A. The percentage yield of gold chloride is 58%
B. The number of mole of Al produced is 40 moles
2Au + 3Cl₂ → 2AuCl₃
Molar mass of Au = 197 g/mol
Mass of Au from the balanced equation = 2 × 197 = 394 g
Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol
Mass of Cl₂ from the balanced equation = 3 × 71 = 213 g
Molar mass of AuCl₃ = 197 + (3×35.5) = 303.5 g/mol
Mass of AuCl₃ from the balanced equation = 2 × 303.5 = 607 g
We'll begin by determining the limiting reactant
From the balanced equation above,
394 g of Au reacted with 213 g of Cl₂
Therefore,
39.4 g of Au will react with = (39.4 × 213) / 394 = 21.3 g of Cl₂
From the above, both Au and Cl₂ are limiting reactant.
Next, we shall determine the theoretical yield.
From the balanced equation above,
394 g of Au reacted to produce 607 g of AuCl₃
Therefore,
39.4 g of Au will react to produce = (39.4 × 607) / 394 = 60.7 g of AuCl₃
Thus, the theoretical yield of AuCl₃ is 60.7 g
Finally, we shall determine the percentage yield.
Actual yield = 35.2 g
Theoretical yield = 60.7 g
Percentage yield =?
Percentage yield = (Actual / Theoretical) × 100
Percentage yield = (35.2 / 60.7) × 100
Percentage yield = 57.99%
Percentage yield ≈ 58%
Balanced equation
2Al₂O₃ → 4Al + 3O₂
From the balanced equation above,
2 moles of Al₂O₃ produced 4 moles of Al
Therefore,
20 moles of Al₂O₃ will produce = (20 × 4) / 2 = 40 moles of Al
Thus, 40 moles of Al were obtained from the reaction.
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