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White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $\mathrm{HCl}$ obtained by the hydrolysis of the product formed by the reaction of $62 \mathrm{~g}$ of white phosphorus with chlorine in the presence of water.
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Reaction of white phsophorus with chlorine and the hydrolyses of the product is given as following :
$$
\begin{gathered}
\mathrm{P}_4+6 \mathrm{Cl}_2 \longrightarrow 4 \mathrm{PCl}_3 \\
\left.\mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl}\right] \times 4 \\
\mathrm{P}_4+6 \mathrm{Cl}_2+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}_3 \mathrm{PO}_3+12 \mathrm{HCl}
\end{gathered}
$$
1 mol. of white phsophorusi.e. $(31 \times 4=124 \mathrm{~g})$ produces 12 mol. of $\mathrm{HCl}$ i.e. $(12 \times 36.5=438.0 \mathrm{~g})$.
$\therefore 62 \mathrm{~g}$ of white phosphorus will produces
$$
\mathrm{HCl}=\frac{438}{124} \times 62=219.0 \mathrm{~g} \mathrm{HCl}
$$
$$
\begin{gathered}
\mathrm{P}_4+6 \mathrm{Cl}_2 \longrightarrow 4 \mathrm{PCl}_3 \\
\left.\mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl}\right] \times 4 \\
\mathrm{P}_4+6 \mathrm{Cl}_2+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}_3 \mathrm{PO}_3+12 \mathrm{HCl}
\end{gathered}
$$
1 mol. of white phsophorusi.e. $(31 \times 4=124 \mathrm{~g})$ produces 12 mol. of $\mathrm{HCl}$ i.e. $(12 \times 36.5=438.0 \mathrm{~g})$.
$\therefore 62 \mathrm{~g}$ of white phosphorus will produces
$$
\mathrm{HCl}=\frac{438}{124} \times 62=219.0 \mathrm{~g} \mathrm{HCl}
$$
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