Question

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

• A. 6 : 5
• B. 5 : 6
• C. 1 : 2
• D. 2 : 1

Answer 1

The correct option is A 6 : 5

$2{KMnO}_4+5{Sn}^{2+}+16H^{+}\to {5Sn}^{4+}+2{Mn}^{2+}+8H_{2}O+2K^{+}$

$\{\left(0.1mol{Sn}^{2+}\right)\times \left(2mol{KMnO}_4/5mol{Sn}^{2+}\right)\}=0.04mol{KMnO}_4$

$K_2{Cr}_2{O}_7+3{Sn}^{2+}+14H^{+}\to 3{Sn}^{4+}+2{Cr}^{3+}+7H_{2}O+2K^{+}$

$\{\left(0.1mol{Sn}^{2+}\right)\times \left(1mol{K}_2{Cr}_2{O}_7/3mol{Sn}^{2+}\right)\}=0.0333mol{K}_2{Cr}_2{O}_7$

$\left(0.04mol{KMnO}_4\right)/\left(0.0333mol{K}_2{Cr}_2{O}_7\right)=\frac{0.12}{0.1}=1.2=\frac{6}{5}$

So, the ratio required to oxidize is $6:5$