Question
White phosphorous reacts with limited chlorine and the product is hydrolysed 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 ?
A.
200 $$g$$
B.
400 $$g$$
C.
219 $$g$$
D.
100 $$g$$
Answer :
219 $$g$$
Solution :
Phosphorus reacts with limited $$C{l_2}$$ as :
$$\eqalign{
& {P_4} + 6C{l_2} \to 4PC{l_3}\,\,\,...\left( {\text{i}} \right) \cr
& PC{l_3}\,{\text{is hydrolysed as :}} \cr
& \underline {PC{l_3} + 3{H_2}O \to {H_3}P{O_3}\left. { + 3HCl} \right] \times 4\,\,\,\,\,\,\,\,\,\,\,\,} ...\left( {{\text{ii}}} \right) \cr
& \underline {{P_4} + 6C{l_2} + 12{H_2}O \to 4{H_3}P{O_3} + 12HCl} \cr} $$
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 $$12 \times 0.5 = 6\,mol$$ of $$HCl$$
Mass of $$HCl = 6 \times 36.5 = 219\,g$$