In the reaction given below:
$x H I + y H N O_{3} ⟶ N O + I_{2} + H_{2} O$

x = 3, y = 2

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x = 2, y = 3

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x = 6, y = 2

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x = 6, y = 1

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Solution

The correct option is C $x = 6, y = 2$
$x H I + y H N O_{3} ⟶ N O + I_{2} + H_{2} O$
$⎡ ⎣ \begin{matrix} 2 I^{-} ⟶ I_{2} + 2 e) \times 3 + 5 (N O_{3}^{-} + 3 e^{-} ⟶ + 2 N O) \times 2 \end{matrix} ⎤ ⎦ \begin{matrix} . . . . . . (1) . . . . . . (2) \end{matrix}$
Adding (1) and (2), we get
$\Rightarrow 6 I^{-} + 2 N O_{3}^{-} ⟶ 2 N O + 3 I_{2}$
$6 H I + 2 H N O_{3} ⟶ 2 N O + 3 I_{2} + 4 H_{2} O$
$\Rightarrow x = 6, y = 2$ .

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