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Question
White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of \[\ce{HCl}\] obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Solution
\[\ce{P4 + 6Cl2 -> 4PCl3}\] .....(i)
\[\ce{PCl3 + 3H2O -> H3PO3 + 3HCl} × 4\]
On adding equation (i) and (ii)
\[\ce{P4 + 6Cl2 + 12H2O -> 4H3PO3 + 12HCl}\]
1 mol of white phosphorus produces 12 mol of \[\ce{HCl}\]
62 g of white phosphorus has been taken which is equivalent to `62/124 = 1/2` mol.
Therefore 6 mol \[\ce{HCl}\] will be formed.
Mass of 6 mol HCl = `6 xx 36.5` = 219.0 g \[\ce{HCl}\].
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