Answer:
Percentage yield of O₂ = 73.4%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2KClO₃ —> 2KCl + 3O₂
Next, we shall determine the mass of KClO₃ that decomposed and the mass of O₂ produced from the balanced equation. This can be obtained as follow:
Molar mass of KClO₃ = 39 + 35.5 + (3×16)
= 39 + 35.5 + 48
= 122.5 g/mol
Mass of KClO₃ from the balanced equation = 2 × 122.5 = 245 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 3 × 32 = 96 g
SUMMARY:
From the balanced equation above,
245 g of KClO₃ decomposed to produce 96 g of O₂.
Next, we shall determine the theoretical yield of O₂. This can be obtained as follow:
From the balanced equation above,
245 g of KClO₃ decomposed to produce 96 g of O₂.
Therefore, 400 g 245 g of KClO₃ will decompose to produce
= (400 × 96)/245 = 156.7 g of O₂.
Thus, the theoretical yield of O₂ is 156.7 g
Finally, we shall determine the percentage yield of O₂. This can be obtained as follow:
Actual yield of O₂ = 115.0 g
Theoretical yield of O₂ = 156.7 g
Percentage yield of O₂ =?
Percentage yield = Actual yield /Theoretical × 100
Percentage yield = 115/156.7 × 100
Percentage yield = 11500/156.7
Percentage yield of O₂ = 73.4%
