Question

What is the volume of oxygen liberated when 49 gm of potassium chlorate is decomposed at ST.P.
I K=49U, CI = 35.5 U, 0 = 16 U 1
2KCIO
2KCIO
+ 302​

Answer 1

Explanation:

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.