The ratio of number of moles of $KMn{O}_4$ and $K_{2}C{r}_2{O}_7$ required to oxidise $0.1$mol $S{n}^{+4}$ in acidic medium :

The correct option is A $6:5$

$\stackrel{+7}{KMn{O}_4}\underrightarrow{acidicmedium}\stackrel{+2}{M{n}^{+2}}$

$\stackrel{+6}{K_{2}C{r}_2{O}_7}\underrightarrow{acidicmedium}\stackrel{+3}{C{r}^{+3}}$

$\stackrel{+4}{S{n}^{+4}}\underrightarrow{reducingagent}\stackrel{+2}{S{n}^{+2}}$

Number of equivalents $=moles\times nf$

$n\left(S{n}^{+4}\right)=0.1$

$nf\left(S{n}^{+4}\right)=2$

Equating number of equivalents of $KMn{O}_4$ and $S{n}^{+4}$

$n_1{nf}_1=n_2{nf}_2$

$0.1\times 2=5n_2{n}_{21}=0.04$

Equating number of equivalents $K_2$ $C{r}_2$ $O_7$ and $S{n}^{+4}$

$n_1{f}_1=n_{2}2f_{2}2$

$0.1\times 2=6n_{22}$

$n_{22}=\frac{0.1}{3}$

Ratio of moles of $KMn{O}_3$ and $K_{2}C{r}_2{O}_7=\frac{0.04}{0.1}=\frac{6}{5}$