Question

what is the volume of oxygen liberated when 49gm of potassium chlorate is decomposed at stp
( k= 49u, cl=35.5u, o=16u)​

Answer 1

Explanation:

ANSWER

The chemical equation for decomposition of KClO

3

can be written as:

2KClO

3

→2KCl+3O

2

245 g of KClO

3

gives 67.2 L of O

2

at STP.

12.25 g of KClO

3

gives

245

67.2

×12.25 =3.36 L.