How many moles of K₂Cr₂O₇ can be reduced by 1 mol of Sn²⁺?

$\begin{array}{l}{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}\\ \left({\mathrm{Sn}}^{2+}\to {\mathrm{Sn}}^{4+}+2{\mathrm{e}}^{-}\right)\times 3\end{array}$

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${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+3{\mathrm{Sn}}^{2+}\to 3{\mathrm{Sn}}^{4+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

It is clear from this equation that 3 mol of ${\mathrm{Sn}}^{2+}$ reduces

$\begin{array}{r}\text{ 1 mol of }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\text{ , hence }1\text{ mol of }{\mathrm{Sn}}^{2+}\text{ will reduce }\frac{1}{3}\text{ mol of }\\ {\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\end{array}$