The unbalanced equation for the reaction of $P_{4} S_{3}$ with nitrate in aqueous acidic medium is given below:.
$P_{4} S_{3} + {NO}_{3}^{-} ⟶ H_{3} {PO}_{4} + {SO}_{4}^{2 -} + NO$
The number of moles of water reauired per mole of $P_{4} S_{3}$ is :

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8/3

answer is B.

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Detailed Solution

The Given reaction can be balanced as,

$\begin{aligned} 3 P_{4} S_{3} + 38 {NO}_{3}^{-} + 8 H_{2} O + 2 {HH}^{+} ⟶ 12 H_{3} {PO}_{4} + 9 {SO}_{4}^{2 -} + 38 NO \end{aligned}$

Here, the number of moles of water required is 8/3.

So, option B is correct.

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