Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{\mathrm{2 }\mathrm{m}\mathrm{o}\mathrm{l}}{2\mathrm{K}\mathrm{O}\mathrm{H}\left(\mathrm{a}\mathrm{q}\right)}+\underset{\mathrm{1 }\mathrm{m}\mathrm{o}\mathrm{l}}{{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\left(\mathrm{a}\mathrm{q}\right)}\to {\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O}+\mathrm{H}\mathrm{e}\mathrm{a}\mathrm{t}$
Thus, maximum heat is liberated if 2 mol of KOH reacts with 1 mol${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$(or if the ratio of the amounts of KOH to${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$is 2 : 1).
Let x = volume of KOH, then 100 mL – x = volume of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$
Number of mmol KOH$=\mathrm{x}\times 1$mmol of KOH
Number of mmol of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4=(100-\mathrm{x})\times 0.5$mmol of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$
Maximum heat is liberated if
$\frac{\mathrm{x }\mathrm{m }\mathrm{m}\mathrm{o}\mathrm{l }\mathrm{o}\mathrm{f }\mathrm{K}\mathrm{O}\mathrm{H}}{(100-\mathrm{x})\times \mathrm{0.5 }\mathrm{m }\mathrm{m}\mathrm{o}\mathrm{l }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}=\frac{2}{1}$
So$\text{x}\text{=}\text{50}\text{mL}$