When water is electrolysed, hydrogen and oxgen gas are produced. If `1.008 g` of `H_(2)` is liberated at cathode, calcualate the mass of `O_(2)` is formed at anode.

Answer

8

Explanation

`:' (w_(1))/(w_(2)) = (E_(1))/(E_(2))`, where `E_(1)` and `E_(2)` are eq. wt. of `H_(2)` and `O_(2)` respectively
`:. (1.008)/(w_(2)) = (1.008)/(8) implies w_(2) = 8 g`