The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product ‘A’. The reaction of 1 mol of ‘A’ with excess of $A g N O_{3}$ in aqueous medium gives mol(s) of Ag.
(Round off to the Nearest Integer)

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6.0

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6.00

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Solution

$P_{4} + 3 N a O H + 3 H_{2} O ⟶ P H_{3} + 3 N a H_{2} P O_{2}$
$P H_{3} + 6 A g N O_{3} ⟶ [A g_{3} P .3 A g N O_{3}] + 3 H N O_{3}$
$[A g_{3} P .3 A g N O_{3}] + 3 H_{2} O ⟶ 6 A g + 3 H N O_{3} + H_{3} P O_{3}$
So, 1 mol of $P H_{3} (A)$ on reaction with excess of aq. $A g N O_{3}$ gives 6 moles of $A g$ .

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The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product ‘A’. The reaction of 1 mol of ‘A’ with excess of AgNO3 in aqueous medium gives mol(s) of Ag. (Round off to the Nearest Integer)

P4+3NaOH+3H2O⟶PH3+3NaH2PO2 PH3+6AgNO3⟶[Ag3P.3AgNO3]+3HNO3 [Ag3P.3AgNO3]+3H2O⟶6Ag+3HNO3+H3PO3 So, 1 mol of PH3(A) on reaction with excess of aq. AgNO3 gives 6 moles of Ag.

The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product ‘A’. The reaction of 1 mol of ‘A’ with excess of $A g N O_{3}$ in aqueous medium gives mol(s) of Ag.
(Round off to the Nearest Integer)