Question

what is the volume of oxygen liberated when 49gm of potassium chlorate is decomposed at s.t.p( k=​

Answer 1

Answer:

suppose

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.

do same as it. you'll get ans.

Answer 2

Answer:

answer is 134.4

Explanation:

672÷245*49

i hope it helps you....