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(B) The reaction of ethanol with $PCl_{3}$ produces ethyl chloride and phosphorous acid $(H_{3}PO_{3})$,which is compound $A$. $3C_{2}H_{5}OH + PCl_{3

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$2$

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(B) The reaction of ethanol with $PCl_{3}$ produces ethyl chloride and phosphorous acid $(H_{3}PO_{3})$,which is compound $A$. $3C_{2}H_{5}OH + PCl_{3} \rightarrow 3C_{2}H_{5}Cl + H_{3}PO_{3} (A)$ Further reaction of $H_{3}PO_{3}$ with $PCl_{3}$ yields pyrophosphorous acid $(H_{4}P_{2}O_{5})$,which is compound $B$. $2H_{3}PO_{3} + PCl_{3} \rightarrow H_{4}P_{2}O_{5} (B) + PCl_{3} + ...$ (The reaction stoichiometry leads to $H_{4}P_{2}O_{5}$). In the structure of pyrophosphorous acid $(H_{4}P_{2}O_{5})$,there are two $P-H$ bonds. Protons attached directly to the phosphorus atom $(P-H)$ are non-ionisable. Therefore,there are $2$ non-ionisable protons in $H_{4}P_{2}O_{5}$.

The number of non-ionisable protons present in the product $B$ obtained from the following reaction is:
$3C_{2}H_{5}OH + PCl_{3} \rightarrow 3C_{2}H_{5}Cl + H_{3}PO_{3} (A)$
$H_{3}PO_{3} + PCl_{3} \rightarrow H_{4}P_{2}O_{5} (B) + HCl$

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The number of non-ionisable protons present in the product $B$ obtained from the following reaction is: $3C_{2}H_{5}OH + PCl_{3} \rightarrow 3C_{2}H_{5}Cl + H_{3}PO_{3} (A)$ $H_{3}PO_{3} + PCl_{3} \rightarrow H_{4}P_{2}O_{5} (B) + HCl$