Question

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2KClO3 mc019-1.jpg 2KCI + 3O2 What is the percent yield of oxygen in this chemical reaction?

Answer 1

The percent yield of oxygen in this chemical reaction is 73.71 %.

Step-by-step explanation:

Experimental yield of oxygen gas = 115.0 g

Theoretical yield:

`2KClO_3rightarrow 2KCI + 3O_2`

Mass of potassium chlorate = 400.0 g

Moles of potassium chlorate =
`(400 g)/(122.5 g/mol)=3.2653 mol`

According to reaction, 2 moles of potassium chlorate gives 3 moles of oxygen gas.

Then 3.2653 mol of potassium chlorate will give:

`(3)/(2)* 3.2653 mol=4.89795 mol`

Mass of oxygen gas :

`32 g/mol* 4.89795 mol=156.73 g`

Percentage yield:

`\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`\%=(115.0 g)/(156.73 g)* 100=73.71\%`

The percent yield of oxygen in this chemical reaction is 73.71 %.

Answer 2

The balanced equation that describes the reaction of heatinf potassium chlorate to produce potassium chloride and oxygen is expressed 2KClO3 = 2KCI + 3O2. For a 400 g potassium chloride, using stoichiometry, the mass oxygen produced is 156.67 grams oxygen. The actual product weighed 115.0 grams. Yield is equal to 115/156.67 or 73.74%