How many hour are required for a current of `3.0` ampere to decompose `18 g` water?

Explanation

`H_(2)O rarr H_(2)+1//2O_(2) [(2H^(+)+2e rarr H^(2)),(O^(2-) rarr (1)/(2)O_(2)+2e)]`
`:.` Eq. of `H_(2)O = (i.t)/(96500)`
Equivalent weight of `H_(2)O = 18//2` as two electrons are used for `1` mole `H_(2)O` to decompose in `H_(2)O` and `O_(2)`.
`:. (18)/(18//2) = (3 xx t)/(96500)`
`t = 64333.3 sec`
`= 1072.2` minute `= 17.87` hr