Question

calculate the mass of potassium chlorate required to librate 6.72l of oxygen at stp molar mass of potassium chlorate is112.5 g/mole

Answer 1

The reaction is as follows:
2KClO₃ → 2KCl + 3O₂
Thus, we can see that
2mol KClO₃ will produce 3 mol O₂
At STP 1 mol O₂ will occupy a volume of 22.4L
Given, 6.72 L of O₂ is liberated
6.72L = 6.72 / 22.4
= 0.3 mol O₂
This will surely require 0.2 mol KClO₃
Hence,
Mass KClO₃ required = 0.2 x 122.5 = 24.5g