Question

`KClO_(3)` on heating decomposes to give `KCl and O_(2)`. What is the volume of `O_(2)` at N.T.P liberated by 0.1 mole of `KClO_(3)` ?

Answer 1

The chemical equation for the decomposition of `KClO_(3)` is
`{:(2KClO_(3),overset("Heat")(rarr),2KCl+3O_(2)),("2 mol",,"3 mol"),(,,(3xx22.4L=67.2L)):}`
2 moles of `KClO_(3)` evolve `O_(2)` at N.T.P = 67.2 L
1 mole of `KClO_(3)` evolve `O_(2)` at N.T.P `= (67.2)/(2)L`
0.1 mole of `KClO_(3)` evolve `O_(1)` at N.T.P `= (67.2)/(2)xx0.1 L =3.36 L`.