Yellow precipitate in the detection of phosphorous when an organic compound is heated with $N a_{2} O_{2}$ and then boiled with conc. $H N O_{3}$ followed by the addition of ammonium molybdate is due to the formation of_________.

(N H_{4})_{3} \cdot P O_{4} \cdot 12 M o O_{3}

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(N H_{4})_{3} \cdot P O_{4} \cdot 6 M o O_{3}

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(N H_{4})_{3} \cdot P O_{4} \cdot 12 M o O_{2}

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(N H_{4})_{3} P O_{4} \cdot 6 M o O_{2}

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Solution

The correct option is A $(N H_{4})_{3} \cdot P O_{4} \cdot 12 M o O_{3}$
$2 P + 3 N a_{2} O_{2} + O_{2} a i r f u s i o n - ------- \to 2 N a_{3} P O_{4}$
$N a_{3} P O_{4} + 12 (N H_{4})_{3} M o O_{4} + 21 H N O_{3} ⟶ (N H_{4})_{3} P O_{4} .12 M o O_{3} c a n a r y y e l l o w + 21 N H_{4} N O_{3} + 12 H_{2} O$
The organic compound is fused with sodium peroxide. The fused mass is then extracted with water. The aqueous solution so obtained is boiled with concentrated nitric acid, and ammonium molybdate solution is added to it.
A yellow solution or precipitate indicates the presence of phosphorus in the organic compound. The yellow precipitate is of ammonium phosphomolybdate

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