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Redox Reactions

The volume of...

Question

The volume of $H N O_{3}$ (having specific gravity of $1.05 g / m L$ containing $8 %$ by mass of $H N O_{3}$ ) required to oxidize iron in $1 g F e S O_{4} \cdot 7 H_{2} O$ in acid medium as $F e^{2 +} + H N O_{3} + H^{+} ⟶ F e^{3 +} + N O + H_{2} O$ is $x \times 10^{- 1}$ . The value of $x$ is :

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Solution

$F e^{2 +} + H N O_{3} + H^{+} ⟶ F e^{3 +} + N O + H_{2} O$
(+5) (+2)
Number of equivalents of $H N O_{3} =$ Number of equivalents of $F e^{+ 2}$
$V \times 1.05 \times \frac{8}{100} \times 21 = \frac{1}{278}$
Therefore, $V = 0.89$ mL.
Hence, volume of $H N O_{3}$ required = $0.89$ mL.

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