Number of moles of $K_{2}C{r}_2{O}_7$  reduced by one mole of $S{n}^{+2}$ ion is

Over all reaction

${\mathrm{Cr}}_2^{+6}{\mathrm{O}}_7^{-2}+14{\mathrm{H}}^{+}+3{\mathrm{Sn}}^{+2}\to 2{\mathrm{Cr}}^{+3}+3{\mathrm{Sn}}^{+4}+7{\mathrm{H}}_2\mathrm{O}$

Thus 3 moles of$S{n}^{+2}$ will reduce 1 moles of $K_{2}C{r}_2{o}_7$ 

$\therefore 1\text{ mole of }S{n}^{+2}\text{ will reduce }\frac{1}{3}\text{ moles }{K}_{2C_2{O}_7}$