$P b + C o n c . H N O_{3}$ gives :

P b (N O_{3})_{2} + N O_{2}

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P b (N O_{3})_{2} + N O

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P b (N O_{3})_{2} + N O_{3}

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P b (N O_{3})_{2} + N_{2} O

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None of these

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Solution

The correct option is B $P b (N O_{3})_{2} + N O_{2}$
$P b + 4 H N O_{3} (c o n c) \to P b (N O_{3})_{2} + 2 H_{2} O + 2 N O_{2}$ .
Hence option A is correct.

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P b (N O_{3})_{2} + N a_{2} S O_{4} ⟶

P b (N O_{3})_{2}

P b {(N O_{3})}_{2}

B a {(N O_{3})}_{2}

19.86

P b (N O_{3})_{2}

[P b = 207, N = 14, O = 16]

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