The percentage purity of cyanogen is:

86.67 %

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76.67 %

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66.67 %

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56.67 %

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Solution

The correct option is C $86.67 %$
$(N H_{4})_{2} C O_{3} \to H_{2} N C O N H_{2} + 2 H_{2} O$

Moles of urea $= \frac{11.56}{96} = 0.12$

$M n O_{4}^{⊝} \equiv (N H_{4})_{2} C_{2} O_{4} (n = 2)$
Milliequivalents of $M n O_{4}^{-} =$ $20 \times 1.6 \times 5 = 160$ milliequivalents

$= \frac{160}{2}$ mmol $= 80$ mmol $= 0.08$ mol

Therefore, moles of $(N H_{4})_{2} C_{2} O_{4} = 0.08$ mol

Let $a$ and $b$ mol of $(C N_{2})$ react in reactions (i) and (ii), respectively.

I. $(C N_{2}) + 4 H_{2} O a = 0.08 m o l \to (N H_{4})_{2} C_{2} O_{4} a = 0.08 m o l$
II. $(C N)_{2} + 2 H_{2} O b = 0.12 m o l \to N H_{2} C O N H_{2} b = 0.12 m o l$

$∴ a = 0.08$ and $b = 0.12$
Total moles of $(C N)_{2} = 0.08 + 0.12 = 0.2$
Weight of $(C N)_{2} = 0.2 \times 52 = 10.4$
$%$ purity of $(C N)_{2} = \frac{10.4}{12} \times 100 = 86.67 %$

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