A sample of hydrazine sulphate `(N_(2)H_(6)SO_(4))` was dissolved in `100 mL` water. `10 mL` of this solution was reacted with excess of `FeCl_(3)` solution and warmed to complete the reaction. Ferrous ions formed were estimated and it required `20 mL` of `M//50 KMnO_(4)` solutions. Estimate the amount of hudrazine sulphate in one litre of solution.
Given `4Fe^(3+)+N_(2)H_(4)rarrN_(2)+4Fe^(2+)+4H^(+)`
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O`

Explanation

The redox changes are
For `FeCl_(3) : Fe^(3+) + e^(-) rarr Fe^(2+)`
For `N_(2)H_(6)SO_(4) : N_(2)^(4-) rarr N_(2) + 4 e^(-)`
For `KMnO_(4) : Mn^(7+) + 5e^(-) rarr Mn^(2+)`
Meq. of `N_(2)H_(6)SO_(4)` in 10 mL solution = Meq. of `FeCl_(3)` reacting with `N_(2)H_(6)SO_(4)` = Meq. of `KMnO_(4)`
Meq. of `N_(2)H_(6)SO_(4)` in 10 mL solution = 20 xx 1/50 xx 5 = 2
`w /(130//4) xx 1000 = 2 ` (Equivalent of `N_(2)H_(6)SO_(4) = 130/4 `)
`w = 2 xx 130 /(4 xx 1000) = 0.065 g`
Weight of `N_(2)H_(6)SO_(4)` in 10 mL = 0.065 g
Thus wt. of `N_(2)H_(6)SO_(4)` in 1000 mL = 6.5 g/L