Sulfuryl chloride $$\left( {S{O_2}C{l_2}} \right)$$   reacts with water to give a mixture of $${H_2}S{O_4}$$  and $$HCl.$$  How many moles of baryta would be required to neutralize the solution formed by adding $$4$$ $$mole$$  of $$S{O_2}C{l_2}$$   to excess of water ?

Solution :
$$\eqalign{ & S{O_2}C{l_2} + 2{H_2}O \to {H_2}S{O_4} + 2HCl \cr & {H_2}S{O_4} + Ba{\left( {OH} \right)_2} \to BaS{O_4} + 2{H_2}O \cr & 2HCl + Ba{\left( {OH} \right)_2} \to BaC{l_2} + 2{H_2}O \cr & {\text{Total}}\,\,moles\,\,{\text{of}}\,Ba{\left( {OH} \right)_2}\,{\text{required}} = 2 \cr} $$