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Variable Oxidation States

Zn reacts wit...

Question

Zn reacts with dilute $H N O_{3}, C o n c . H N O_{3}$ and very dilute $H N O_{3}$ . The oxidation number of nitrogen changes from $H N O_{3}$ are $a, b, c$ in product $X, Y, Z$ respectively. What is the value of $(a + b + c)$ is
(a) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + " X " + H_{2} O$
dilute ( $20 %$ )
(b) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + " Y " + H_{2} O$
Conc ( $70 %$ )
(a) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + " Z " + H_{2} O$
very dilute ( $6 %$ )

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Solution

(a) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + N_{2} O (X) + H_{2} O$
dilute ( $20 %$ )
(b) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + N O_{2} (Y) + H_{2} O$
Conc ( $70 %$ )
(a) $Z n + H N O_{3} \to Z n (N O_{3})_{2} + N H_{4} N O_{3} (Z) + H_{2} O$
very dilute ( $6 %$ )

$Z = (- 3) N H_{4}^{+} N O_{3}^{-} \to - 3$
$X = N_{2} O \to + 1$
$Y = N O_{2} \to + 4$
$∴ a + b + c = 5 - 3 = 2$

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