Question

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? use percent yield equals startfraction actual yield over theoretical yield endfraction times 100.. 69.63% 73.40% 90.82% 136.2%

Answer 1

The percentage yield of oxygen obtained from the reaction, given that 400 g of potassium chlorate was heated is 73.4%

Balanced equation

2KClO₃ —> 2KCl + 3O₂

Molar mass of KClO₃ = 122.5 g/mole

Mass of KClO₃ from the balanced equation = 2 × 122.5 = 245 g

Molar mass of O₂ = 32 g/mole

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

How to determine the theoretical yield

From the balanced equation above,

245 g of KClO₃ decomposed to produce 96 g of O₂

Therefore,

400 g of KClO₃ will decompose to produce = (400 × 96) / 245 = 156.7 g of O₂

How to determine the percentage yield

• Actual yield of O₂ = 115 g
• Theoretical yield of O₂ = 156.7 g
• Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (115 / 156.7) × 100

Percentage yield = 73.4%

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Answer 2

Answer:

B. 73.40%

Explanation:

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