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 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%

Answer 2

Answer:

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

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 = $\frac{400 g}{122.5 g/mol}=3.2653 mol$

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

Then 3.2653 mol of potassium chlorate will give:

$\frac{3}{2}\times 3.2653 mol=4.89795 mol$

Mass of oxygen gas :

$32 g/mol\times 4.89795 mol=156.73 g$

Percentage yield:

$\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%=\frac{115.0 g}{156.73 g}\times 100=73.71\%$

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