Ratio of stoichiometric coefficients of ${\mathrm{CrCl}}_3$ to ${{\mathrm{MnO}}_4}^{-}$ in balanced chemical reaction in acidic medium is ${{\mathrm{MnO}}_4}^{-} + {\mathrm{CrCl}}_3 \to {\mathrm{MnCl}}_2+ {\mathrm{Cr}}_2{\mathrm{O}}_7{ }^{2-}+ {\mathrm{Cl}}^{-}$

$$\begin{gathered} \stackrel{+7}{{{\mathrm{MnO}}_4}^{-}} + \stackrel{+3}{{\mathrm{CrCl}}_3} \to \stackrel{+2}{{\mathrm{MnCl}}_2}+ \stackrel{+6}{{\mathrm{Cr}}_2{\mathrm{O}}_7{ }^{2-}}+ {\mathrm{Cl}}^{-} \\ \mathrm{RHR} \mathrm{OHR} \\ {\mathrm{Mn}}^{+7}+ 5{\mathrm{e}}^{-} \to {\mathrm{Mn}}^{+2} ....\left(\mathrm{i}\right)\times 6 2{\mathrm{Cr}}^{+3} \to {{\mathrm{Cr}}_2}^{+6}+ 6{\mathrm{e}}^{-} .... \left(\mathrm{ii}\right) \times 5 \\ 6{{\mathrm{MnO}}_4}^{-} + 10 {\mathrm{CrCl}}_3 + 11{\mathrm{H}}_2\mathrm{O} \to 6 {\mathrm{MnCl}}_2 + 5{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2} + 18 {\mathrm{Cl}}^{-} + 22{\mathrm{H}}^{+} \\ \frac{\mathrm{stoichiometric} \mathrm{coefficient} \mathrm{of} {\mathrm{CrCl}}_3 }{\mathrm{stoichiometric} \mathrm{coefficient} \mathrm{of} {\mathrm{MnO}}_4}=\frac{10}{6}=\frac{5}{3} \mathrm{Ans} \end{gathered}$$