Question

A sample of hydrazine sulphate $\left(N_2{H}_{6}S{O}_4\right)$ was dissolved in $100$ mL of water and $10$ mL of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required $20$ mL of $\frac{M}{50}$ potassium permanganate solution. Estimate the amount of hydrazine sulphate in $1$ L of the solution. Reactions are given below:
$4F{e}^{3+}+N_2{H}_4\to N_2+4F{e}^{2+}+4H^{⨁}$.
$Mn{O}_4^{⊝}+5F{e}^{2+}+8H^{⨁}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

• A. $1.91$ g
• B. $3.82$ g
• C. $2.71$ g
• D. $1$ g

Answer 1

The correct option is A $1.91$ g

The reactions are as follows:

$4F{e}^{3+}+N_2{H}_4\to N_2+4F{e}^{2+}+4H^{⨁}$.
$Mn{O}_4^{⊝}+5F{e}^{2+}+8H^{⨁}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

$1$ mol $N_2{H}_4\equiv 4$ mol $F{e}^{3+}\equiv \frac{4}{5}$ mol $KMn{O}_4$
$20$ mL of $\frac{M}{50}$ potassium permanganate solution corresponds to $\frac{20\times 1}{1000\times 50}=4\times {10}^{-4}$ moles.
The number of moles of hydrazine are $\frac{4}{5}\times 4\times {10}^{-4}=3.2\times {10}^{-4}$ moles.
The molar mass of hydrazine is $32$ g/mol.
$3.2\times {10}^{-4}$ moles of hydrazine corresponds to $3.2\times {10}^{-4}\times 32=0.01024$ g
This is the amount of hydrazine present in $10$ mL of solution.
The amount of hydrazine present in $1$ L of solution will be $100\times 0.01=1$ g.