$H_{2} O_{2}$ turns blackened lead paintings to white colour. In this reaction it oxidizes $P b S$ to $P b S O_{4}$ . The number of moles of $H_{2} O_{2}$ needed to oxidize $0.1$ moles of $P b S$ is:

1 mole

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0.1 mole

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0.5 mole

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0.4 mole

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Solution

The correct option is D 0.4 mole
The balanced equation is:
$P b S + 4 H_{2} O_{2} \to P b S O_{4} + 4 H_{2} O$ .

The stiochiometric coefficient of $P b S$ is $1$ while that of $H_{2} O_{2}$ is $4$ in the balanced reaction.

So, 0.1 mol will be oxidised by 0.4 mol of $H_{2} O_{2}$

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