For testing nitrite ions we used two different reagents. First one is urea and another one is sulphamic acid. In both the reactions we get gases $A$ and $B$ which are:

N_{2}

and

N O_{2}

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N_{2}

and

N_{2}

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N O_{2}

and

N_{2}

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N O_{2}

and

N_{2} O

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Solution

The correct option is B $N_{2}$ and $N_{2}$
To find: The gases obtained by the reaction of nitrite ion and urea and nitrite ion and sulphamic acid.
Reaction of nitrite ion with urea yields carbon dioxide and nitrogen gas.
$H_{2} N C O N H_{2} (u r e a) + 2 H N O_{2} (n i t r i t e i o n) \to C O_{2} + 2 N_{2} + 3 H_{2} O (R e a c t i o n w i t h u r e a)$
$H N O_{2} (n i t r i t e i o n) + H_{3} N S O_{3} (s u l p h a m i c a c i d) \to H_{2} S O_{4} + N_{2} + H_{2} O$

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Z n + c o n c . H N O_{3} (h o t) \to Z n (N O_{3})_{2} + X + H_{2} O

Z n + d i l . H N O_{3} (cold) \to Z n (N O_{3})_{2} + Y + H_{2} O

N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

- \frac{d [{N O}_{2}]}{d t} = K_{2} [{N O}_{2}]

- \frac{d [O_{2}]}{d t} = K_{3} [O_{2}]

K_{1}, K_{2}, K_{3}

A g N O_{3} (s)

N H_{4} N O_{2}

N H_{4} N O_{3}

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