Solution

Ionic form of reaction is:
$$Cr_2O_7^{2-} + Sn^{+2} +H^+ \rightarrow Sn^{+4}+ Cr^{+3}+H_2O$$

Step 1: Assign the oxidation state using oxidation number of $$O=-2$$.
we get:
$$\overset {+6}{Cr_2}O_7^{2-} + \overset {+2}{Sn^{2+}} + H^+ \rightarrow \overset {+4}{Sn^{4+}} + \overset {+3}{Cr^{3+}} + H_2O$$

Reduction half-reaction:
$$\overset {+6}{Cr_2}O_7^{2-} \rightarrow 2\overset {+3}{Cr^{3+}}$$: Gain of 6 electrons

Oxidation half reaction:
$$\overset {+2}{Sn^{2+}} \rightarrow \overset {+4}{Sn^{4+}}$$: Loss of 2 electrons

Step 2: Equalise the number of electrons as:

Oxidation half reaction:
$$3\overset {+2}{Sn^{2+}} \rightarrow 3\overset {+4}{Sn^{4+}}$$

Step 3: balance O atoms by adding $$H_2O$$ and then H by $$H^+$$
$$\overset {+6}{Cr_2}O_7^{2-} +14H^+ \rightarrow 2\overset {+3}{Cr^{3+}}+7H_2O$$

Step 4: overall reaction:
$$\overset {+6}{Cr_2}O_7^{2-} +14H^++3\overset {+2}{Sn^{2+}} \rightarrow 2\overset {+3}{Cr^{3+}}+3\overset {+4}{Sn^{4+}}+7H_2O$$

Thus 3 mole of $$Sn^{2+}$$will reduce 1 moles of $$K_2Cr_2O_7$$.

Therefore, 1 mole of $$Sn^{2+}$$will reduce $$\frac { 1 }{ 3 }$$ moles of $$K_2Cr_2O_7$$.