White phosphorous reacts with limited chlorine and the product is hydrolyzed in the presence of water. What would be 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?

200 g

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400 g

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219 g

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100 g

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Solution

The correct option is D 219 g
Phosphorus reacts with limited $C l_{2}$ as:
$P_{4}$ + 6 $C l_{2}$ $\to$ 4 $P C l_{3}$ ...........(i)
$P C l_{3}$ is hydrolysed as :
$P C l_{3}$ + 3 $H_{2} O$ $\to$ $H_{3} P O_{3}$ + 3HCl ] x 4... (ii)
$P_{4}$ + 6 $C l_{2}$ + 12 $H_{2} O$ $\to$ 4 $H_{3} P O_{3}$ + 12HCl
Moles of white $P_{4}$ = $\frac{62}{124}$ = 0.5 mol
1 mol of white $P_{4}$ produces = 12 mol of HCl
0.5 mol of white $P_{4}$ will produce 12 x 0.5 = 6 mol of HCl
Mass of HCl = 6 x 36.5 = 219 g

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White phosphorous reacts with limited chlorine and the product is hydrolyzed in the presence of water. What would be 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?

White phosphorous reacts with limited chlorine and the product is hydrolyzed in the presence of water. What would be 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?