How many grams of O2 can be prepared from the decomposition of 4.50 grams of potassium chlorate?

Question

How many grams of O2 can be prepared from the decomposition of 4.50 grams of potassium chlorate?

Answer 1

Potassium chlorate is KClO3 and the decomposition looks like this...

2KClO3 ==> 2KCl + 3O2

4.50 g KClO₃ x 1 mole/122.55 g = 0.0367 moles

0.0367 moles KClO₃ x 3 moles O₂/2 moles KClO₃ = 0.0245 moles O₂ (see mo

le ratio in balanced equation)

mass of O₂ = 0.0245 moles O₂ x 32 g/mole = 0.784 g O₂ formed