By passing electric current,${\mathrm{NaClO}}_3$ in to NaClO₄ according to the following equation ${\mathrm{NaClO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NaClO}}_4+{\mathrm{H}}_2$ How many moles of ${\mathrm{NaClO}}_4$ will be formed when three faradays of charge is passed through NaClO₃ ? 

$$\begin{gathered} \mathrm{Na}\stackrel{+5}{{\mathrm{ClO}}_3}+\hspace{0.17em}{\mathrm{H}}_2\mathrm{O}\to \mathrm{Na}\stackrel{+7}{{\mathrm{ClO}}_4}+\hspace{0.17em}{\mathrm{H}}_2 \\ {\mathrm{Cl}}^{+5 }\to {\mathrm{Cl}}^{+7}+\hspace{0.17em}2{\mathrm{e}}^{-} \\ 2 \mathrm{mol} \mathrm{of} \mathrm{electrons} =2 \mathrm{Faraday} \\ \mathrm{It} \mathrm{is} \mathrm{clear} \mathrm{that} \mathrm{for} \mathrm{formation} \mathrm{of} 1\mathrm{mol} \mathrm{of} {\mathrm{NaClO}}_4 , \mathrm{required} 2\mathrm{F}\hspace{0.17em} \\ 2\mathrm{F} = 1\mathrm{mol} {\mathrm{NaClO}}_4 \\ 1\mathrm{F}\hspace{0.17em}= \frac{1}{2} \mathrm{mol} {\mathrm{NaClO}}_4 \\ \mathrm{When} 3\mathrm{F} \mathrm{of} \mathrm{charge} \mathrm{passed} \\ = 3\hspace{0.17em}\times \frac{1}{2} \mathrm{mol} \mathrm{of} {\mathrm{NaClO}}_{4 } \\ =1.5 \mathrm{mol} \mathrm{of} {\mathrm{NaClO}}_4 \mathrm{formed} \end{gathered}$$