When $2$ mol of lead nitrate is heated, then:

2

mol of

P b O_{2}

gas are produced

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4

mol of

N O_{2}

gas are produced

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2

mol of

O_{2}

gas are produced

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All of these

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Solution

The correct option is B $4$ mol of $N O_{2}$ gas are produced
The reaction involved is:
$2 P b (N O_{3})_{2} Δ - \to 2 P b O + 4 N O_{2} + O_{2}$

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