One mole of ${\mathrm{CaOCl}}_2$ is dissolved in water and excess of KI added. Hypo $\left({\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\right)$ required to react with the oxidized part completely is

${\mathrm{CaOCl}}_2 + {\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to } \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 + {\mathrm{Cl}}_2$

1 mol                                                1 mol

${\mathrm{Cl}}_2 + 2\mathrm{KI} \stackrel{ }{\to } 2\mathrm{KCl} + {\mathrm{I}}_2$

                                         1 mole

$$\begin{gathered} {\mathrm{I}}_2 + \mathrm{KI} \rightleftharpoons {\mathrm{KI}}_3 \\ {\mathrm{I}}_2 \left(\mathrm{in} \mathrm{KI}\right) + 2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3 \stackrel{ }{\to } 2\mathrm{NaI} + {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6 \end{gathered}$$