Question

What volume of oxygen atoms S.T.P. can be produced by 6.125 g of potassium chlorate

according to the example

2KClO3 2KCl + 3O2 [M.W. of KClO3 = 122.5]​

Answer 1

Answer:

The number of moles of KClO

3

is 6.125 g

So,

122.5

6.125

= 0.05 moles.

Two molecules of KClO3

requires 3 molecules of O2

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