A sample of hydrazine sulphate ($ ext{N}_2 ext{H}_6 ext{SO}_4$) was dissolved in $100 ext{ mL}$

Question

A sample of hydrazine sulphate ($	ext{N}_2	ext{H}_6	ext{SO}_4$) was dissolved in $100	ext{ mL}$ of water. $10	ext{ mL}$ of this solution was reacted with an excess of $	ext{FeCl}_3$ solution and warmed. The $	ext{Fe}^{2+}$ formed required $20	ext{ mL}$ of $\frac{	ext{M}}{50}	ext{ KMnO}_4$ solution for complete oxidation. If the amount of hydrazine sulphate in one litre of the original solution is $y	ext{ g}$, find the value of $10y$. (Given molar mass of $	ext{N}_2	ext{H}_6	ext{SO}_4 = 130	ext{ g/mol}$)

Professor PC Thomas Classes · Answer

Moles of $	ext{KMnO}_4 = 20 	imes 0.02 = 0.4	ext{ mmol}$. Equivalents of $	ext{KMnO}_4 = 0.4 	imes 5 = 2.0	ext{ meq}$. Thus, $	ext{Fe}^{2+}$ formed = $2.0	ext{ mmol}$. The reaction is $4	ext{Fe}^{3+} + 	ext{N}_2	ext{H}_4 ightarrow 	ext{N}_2 + 4	ext{Fe}^{2+} + 4	ext{H}^+$. Moles of $	ext{N}_2	ext{H}_4$ in $10	ext{ mL} = \frac{2.0}{4} = 0.5	ext{ mmol}$. In $100	ext{ mL}$, moles = $5.0	ext{ mmol} = 0.005	ext{ mol}$. Concentration in $1	ext{ L}$ = $0.05	ext{ M}$. Mass in $1	ext{ L}$ ($y$) = $0.05 	imes 130 = 6.5	ext{ g}$. Therefore, $10y = 65$.