A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2KClO3 Right arrow. 2KCI 3O2 What is the percent yield of oxygen in this chemical reaction

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%

answer: mole of the gas

answer: mole of the gasexplanation: