Question

Calculate the mass of potassium chlorate required to liberate 6.72dm3 of oxygen at stp molar mass of potassium chlorate is 122.5g/mol.

Answer 1

Start with the balanced chemical equation and then use the unit-factor method. Also, everyone knows that 1 mol of any ideal gas at STP occupies a volume of 22.4L. Use that as a conversion factor. 

2KClO3(s) --> 2KCl(s) + 3O2(g) 
???g ............ .................. 6.72L (STP) 

6.72 L x (1 mol O2 / 22.4L) x (2 mol KClO3 / 3 mol O2) x (122.5g KClO3 / 1 mol KClO3) = 24.5g KClO3