Question

Calculate the mass and volume of oxygen released when 29.4g pottasium chlorate decomposes to form pottasium chloride and oxygen

Answer 1

2KClO3 ==> 2KCl + 3O2
moles KClO3 = g/molar mass = 29.4/122.6 = approx 0.24
2 mols KClO3 will produce 3 mols O2; thereofre, 0.24 mols KClO3 will produce 0.24 x 3/2 = approx 0.36
1 mol O2 will occupy a volume of 22.4 L @ STP so 0.36 mols will have a volume of .......?
mass O2 = moles x molar mass = ?

Answer 2

2KClO3(s)>2KCl(s)+3O2(g)
RFM of KClO3 =39+35.5+3×16 =122.5
moles of KClO3 =29.4/122.5 =0.33moles
mole ratio of KClO3 to O2 is 2:3
moles of O2 =3/2×0.33= 0.495moles

mass=moles×molar mass
=0.495×16
=7.92g

1 mole at STP=22.4dm^3
0.495 moles=??
=0.495×22.4
=11.088dm^3