A sample of hydrazine sulphate `(N_2H_6SO_4)` was dissolved in 100 " mL of " water. 10 " mL of " this solution was reacted with excess of `FeCl_3` solution and warmed to complete the reaction. `Fe^(2+)` ions formed were estimated and they required 20 " mL of " `(M)/(50)KMnO_4` Solution. Estimate the amount of hydrazine sulphate in 1 L of solution.
Reactions:
`4Fe^(3+)+N_2H_4toN_2+4Fe^(2+)+4H^(o+)`
`MnO_4^(ɵ)+5Fe^(2+)+8He^(o+)toMn^(2+)5Fe^(3+)+4H_2O`

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