The reaction of white phosphorus on boiling w | vedclass

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Question

(C) The reaction of white phosphorus $(P_4)$ with alkali $(NaOH)$ is a disproportionation reaction:$P_4 + 3 OH^{-} + 3 H_2O \rightarrow PH_3 + 3 H_2PO

Vedclass · Accepted Answer

$4$

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(C) The reaction of white phosphorus $(P_4)$ with alkali $(NaOH)$ is a disproportionation reaction:$P_4 + 3 OH^{-} + 3 H_2O \rightarrow PH_3 + 3 H_2PO_2^{-}$Here,the product '$A$' is the hypophosphite ion $(H_2PO_2^{-})$.The reaction of $1 \ mol$ of $H_2PO_2^{-}$ with excess $AgNO_3$ is a redox reaction:$H_2PO_2^{-} + 4 Ag^{+} + 2 H_2O \rightarrow 4 Ag + H_3PO_4 + 3 H^{+}$From the stoichiometry of the balanced equation,$1 \ mol$ of $H_2PO_2^{-}$ reacts to produce $4 \ mol$ of $Ag$.

The reaction of white phosphorus on boiling with alkali in an inert atmosphere results in the formation of product '$A$'. The reaction of $1 \ mol$ of '$A$' with an excess of $AgNO_3$ in an aqueous medium gives ............... $mol(s)$ of $Ag$. (Round off to the Nearest Integer).

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