Question

12. Using the Figure 13.19 in SA answer the following: 0.8 moles of potassium chlorate are added to 300 ml of boiling water. What is the mass of precipitate formed after the solution is cooled down to 20 oC?

Answer 1

Potassium chlorate when added to boiling water it decomposes into potassium chloride and oxygen gas.

$2 KClO3 \rightarrow 2KCl + O2$

As per balanced equation when 2 moles of potassium chlorate is heated it gives 2 moles of potassium chloride and 1 mole of oxygen gas.

When 0.8 moles of potassium chlorate is added to boiling water it decomposes into 0.8 moles of potassium chloride and 0.4 moles of oxygen gas.

In boiling water potassium chloride will be present in dissolved state and when the solution is cooled to 20^oC it precipitates.

So, the number of moles of potassium chlorate that precipitates from solution = 0.8 moles

Molar mass of potassium chlorate (KCl) = 39 + 35.5 = 78.5

The mass of potassium chlorate precipitated = 0.8 x 78.5 = 62.8 g