The decomposition of potassium chlorate yields oxygen gas. If the yield is 65%, how many grams of KClO3 are needed to produce 47.5L of O2?

Question

chemistry problem

Answer 1

The following answer assumes conditions of STP, otherwise, one cannot answer the question without knowing the  temperature and the pressure.  

 

2KClO3 ==> 2KCl + 3O2

At STP 47.5 L x 1 mol/22.4 L = 2.12 moles O2 formed

2.12 moles O2 x 2 moles KClO3/3 moles O2 = 1.41 moles KClO3 needed

At 65% efficiency, one needs 1.41mol/0.65 = 2.17 moles KClO3

mass KClO3 needed = 2.17 moles x 123 g/mol = 268 g