To `50 L` of `0.2 N NaOH, 5 L` of `1 N HCl` and `15 L` of `0.1 NFeCl_(3)` solution are added. What weight of `Fe_(2) O_(3)` can be obtained from the precipitate? Also report the nomality of `NaOH` left in the resultant solution.

Explanation

Equivalent of `NaOH = 50 xx 0.2 = 10`
Equivalent of `HCl = 5 xx 1 = 5`
`:.` Equivalent of `NaOH` left after reaction with `HCl`
`= 10 - 5 = 5`
Also `NaOH` reacts with `FeCl_(3)` to give `Fe (OH)_(3)` which on ignition gives `Fe_(2) O_(3)`.
`:.` Equivalent of `NaOH` used fro `FeCl_(3)`
= Equvalent of `Fe(OH)_(3)`
Equivalent of `Fe_(2) O_(3)`
`= 15 xx 0.1 = 1.5`
`:.` Equivalent of `NaOH` left finally `= 5 - 1.5 = 3.5`
`N_(NaOH)` left `= (3.5)/(70) = 0.05 N`
`:.` Total volume `= 70 L`
Also, equivalent of `Fe_(2) O_(3) = 1.5 (Mw of Fe_(2)O_(3) = 160)`
`:. (W)/(Mw//6) = 1.5`
`:. W_(Fe_(2)O_(3)) = (1.5 xx 160)/(6) = 40 g`