Question

kclo3 on heating gives kcl and O2 .What is the volume of O2 liberated by 0.1 mole of Kclo3.
​

Answer 1

Answer:

The chemical equation for the decomposition of KClO3 is

2KClO32 mol−→−Heat2KCl+3O23 mol(3×22.4L=67.2L)

2 moles of KClO3 evolve O2 at N.T.P = 67.2 L

1 mole of KClO3 evolve O2 at N.T.P =67.22L

0.1 mole of KClO3 evolve O1 at N.T.P =67.22×0.1L=3.36L

mark me as brainliest

Answer 2

3.36 Litres

The decomposition of KClO3 on heating is called as thermal decomposition reaction.

The Chemical Equation showing this reaction:

2KClO3 ----heating---> 2KCl   + 3O2

2 mol                                           3 mol

Calculating how much 2 moles of KClO3 evolve Oxygen at N.T.P

= 3 x 22.4 L

= 67.2 Litres

Calculating how much 1 mole of KClO3 evolve Oxygen at N.T.P:

For this we take 2 to the other side of the equation

= 67.2 / 2 Litres

= 33.6 Litres

Now,

Calculating how much 0.1 mole of KClO3 evolve Oxygen at N.T.P

= 67.2 / 2 x 0.1 L

= 3.36 Litres

Therefore, the volume of O2 liberated by 0.1 mole of KClO3 is 3.36 Litres.