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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. |
Answer» 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` |
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