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${P b}_{3} O_{4} + 8 H C l ⟶ 3 {P b C l}_{2} + {C l}_{2} + 4 H_{2} O$
In the above reaction the number of moles of $H C l$ are oxidized by one mole of ${P b O}_{2}$

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$3$

$2$

$8$

$1$

answer is D.

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Detailed Solution

Usually ${P b}_{3} O_{4}$ exists as ${P b O}_{2} . 2 P b O$

The oxidation state of chlorine in $H C l$ is $- 1$ , in ${P b C l}_{2}$ is $- 1$ and in ${C l}_{2}$ is $0$

So the given reaction can be return as

${P b O}_{2} \cdot 2 P b O + 6 H C l + 2 H C l ⟶ 3 {P b C l}_{2} + {C l}_{2} + 4 H_{2} O$

The number of moles of $H C l$ oxidized are $2$

Hence the correct option is D.

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