Question

What volume of oxygen at STP can be produced by 6.125g of potassium chlorate according to the reaction 2KClO3----------- 2KCl + 3O2

Answer 1

no of moles of KCIO3=6.125/122.5

=0.05 moles

2molecules of KCLO3 require 3 molecules of 02

number of moles of O2=3*0.05/2

=0.075 moles

volume of 1 mole =22.4L

volume of O2=22.480.075

=1.68 L

Answer 2

Atomic mass of K=39g

Atomic mass of Cl=35.5g

Atomic mass of O=16g

Mass of 2KClO3=2(39+35.5+(16*3))

                          =2(39+35.5+48)

                          =2*122.5

                          =245g

Mass of 3O2=3*16*2

                    =96g

245g of potassium chlorate at STP gives 96g of oxygen

Therefore, 6.125g of potassium chlorate at STP gives (96/245)*6.125=2.4g of oxygen