What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g) according to the equation above?

Question

Chemistry

Humphrey

28 July, 23:48

What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g) according to the equation above?

+3

Answers (1)

Know the Answer?

Anya Shah · Answer 1

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ → 2KCl + 3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1.3) / 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol. 32 g / 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield). 100 =

(3.2g / 4.8g). 100 = 66.7 %