An aqueous solution of $F e C l_{3}$ obtained when an itched copper circuit board was washed required 30 mL of 0.1 M hydrazine hydrochloride $[N H_{2} N H_{2} . H C l]$ for its reduction to $F e^{2 +}$ . During this process hydrazine gets converted to N₂. How many mL of $\frac{4}{3} M$ nitric acid is required for the re-oxidation of $F e^{2 +}$ to $F e^{3 +}$ ?
Given: Nitric acid gets reduced to NO during this process.

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answer is 3.

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Detailed Solution

Number of m. equiv of $F e^{3 +}$ = no. of m. equiv. of hydrazine
= no. of m. equiv. of $F e^{2 +}$
= no. of m. equiv. of nitric acid
$\frac{4}{3} \times 3 \times V =0.1 \times 30 \times 4 \Rightarrow V = 3 m L$

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