One mole of $\mathrm{P}_{4}$ reacts with 8 moles of $\mathrm{SOCl}_{2}$ to give 4 moles of $\mathrm{A}, x$ mole of $\mathrm{SO}_{2}$ and 2 moles of $\mathrm{B} . \mathrm{A}, \mathrm{B}$ and $x$ respectively are :
Solution
$$
\mathrm{P}_4+8 \mathrm{SOCl}_2 \longrightarrow 4 \mathrm{PCl}_3+4 \mathrm{SO}_2+2 \mathrm{~S}_2 \mathrm{Cl}_2
$$
<br/><br/>Hence, $\mathrm{A}=\mathrm{PCl}_3, \mathrm{x}=4, \mathrm{~B}=\mathrm{S}_2 \mathrm{Cl}_2$
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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