Consider reaction: $P_{4} O_{10} (s) + H_{2} O (l) ⟶ H_{3} P O_{4} (a q)$
Moles of $H_{2} O$ required to balance equation will be:

2

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4

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6

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8

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Solution

The correct option is D $6$
Using oxidation number method balancing $p h o s p h o r u s .$
$P_{4} O_{10} ⟶ 4 H_{3} P O_{4}$
Balancing oxygen and hydrogen, $6 H_{2} O + P_{4} O_{10} ⟶ 4 H_{3} P O_{4}$
$∴$ Moles of $H_{2} O$ required $= 6.$

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