If 267 mL of O2, measured at STP, are obtained by the decomposition of the KClO3 in a 3.33 g mixture of KCl and KClO3, 2KClO3(s) --> 2KCl(s) + 3O2

Question

What is the percent by mass of KClO3 in the mixture?

Answer 1

Given that 1 mole of any gas occupies 22.4 litres at STP you can calculate the number of moles of O2 in 267 ml, (F moles)

2KClO3(s) --> 2KCl(s) + 3O2

tells you that 3 moles of O2 came from 2 moles of KClO3, therefore you can work out how many moles of KClO3 would be needed to yield F moles of O2

number of moles of KClO3
=Fx2/3

Calcluate the molar mass of KClO3, M.

Thus the mass of KClO3 needed is

(Fx2xM/3) g

Hence the percentage of KClO3 in the mixture is

Fx2xMx100%/(3x3.33)