White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $HCl$ obtained by the hydrolysis of the product formed by the reaction of $62 \ g$ of white phosphorus with chlorine in the presence of water.

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(219.0 G) The reaction of white phosphorus $(P_4)$ with chlorine $(Cl_2)$ followed by hydrolysis is given by the equation:
$P_4 + 6 Cl_2 + 12 H_2O \rightarrow 4 H_3PO_3 + 12 HCl$
From the stoichiometry,$1 \ mol$ of $P_4$ produces $12 \ mol$ of $HCl$.
Molar mass of $P_4 = 4 \times 31 = 124 \ g/mol$.
Moles of $P_4 = \frac{62 \ g}{124 \ g/mol} = 0.5 \ mol$.
Therefore,moles of $HCl$ produced $= 12 \times 0.5 \ mol = 6 \ mol$.
Molar mass of $HCl = 1 + 35.5 = 36.5 \ g/mol$.
Mass of $HCl = 6 \ mol \times 36.5 \ g/mol = 219.0 \ g$.

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