Hydroxylamine reduces `Fe^(3+)` accoeding to the following reaction:
`2NH_2OH+4Fe^(3+)toH_2O+4Fe^(2+)+4H^(o+)+N_2O`
`Fe^(2+)` produced is is estimated by titration with `KMnO_4` solution A 10 mL sample of `NH_2OH` is diluted to 1000 mL. 50 " mL of " this diluted sample is boiled with excess of Fe (III) solution. The resulting solution required 12 " mL of " 0.02 M `KMnO_4` for complete oxidation. Determine the strength of `NH_2OH`.

Explanation

`2NH_(2)OH+4Fe^(3+)toN_(2)O+4Fe^(2+)+H_(2)O+4H^(o+)`
`MnO_(4)^(ɵ)+5Fe^(2+)+8H^(o+)toMn^(2+)+4H_(2)O`
12 " mL of " 0.02 of M `KMnO_(4)=12xx0.2xx5mEq` of `Fe(2+)`
`=1.2m" " mol of ""Fe^(2+)`
`=1.2xx10^(-3)xx56g` of `Fe^(2+)`
`4xx56g` of `Fe^(2+)=2xx33g` of `NH_(2)OH`
`0.067g` of `Fe^(2+)=(2xx33xx0.672)/(4xx56)`
`=0.0198g` of `NH_(2)OH`
0.0198 g of `NH_(2)OH` is present in 50 " mL of " solution
`therefore` 1000 " mL of " solution will contain `=(0.0198xx1000)/(50)`
`=0.396g` of `NH_(2)OH`
This much dilute solution is obtained from 10 " mL of " the original sample.
Hence strength `=(0.396xx1000)/(10)=39.6gL^(-1)`
Alternatively: `undersetunderset(2x=-2)(2x+6-4=0)(2NH_2OH)toundersetunderset(2x=2)(2x-2=0)(N_2O+4e^(-))(x=(4)/(2)=2)`
" Eq of "`NH_2OH=2xx"moles of "NH_2OH(Ew=(33)/(2))`
`12mL" of "0.02 M KMnO_4-=12xx0.02xx5mEq" of "Fe^(2+)`
`-=1.2mEq" of "NH_2OH`
`-=1.2xx10^(-3)xx(33)/(2)g" of "NH_2OH`
`-=0.0198 g" of "NH_2OH`
After that, follow the above method.