What happens when
(i) Na metal is dropped in water ?
(ii) Na metal is heated in free supply of air ?
(iii) Sodium peroxide `(Na_(2)O_(2))` dissolved in water ?

Explanation

(i) `2Na(s)+2H_(2)O(l)to2NaOH(aq)+H_(2)(g)`
This is an exothermic reaction and due to liberated heat, `H_(2)(g)` easily catches fire.
(ii) `Na_(2)O_(2)` along with a small amount of `Na_(2)O` is formed.
`4Na(s)+O_(2)(g) to 2Na_(2)O(s)`
`Na_(2)O(s)+1//2O_(2)(s)toNa_(2)O_(2)(s)`
(iii) `H_(2)O_(2)` is formed in this reaction like
`Na_(2)O_(2)(s)+2H_(2)O(l)to2NaOH(aq)+H_(2)O_(2)(l)`