Question

2 moles of $FeSO_{4}$ in acid medium are oxidized by x moles of $KMnO_{4}$ , whereas 2 moles of $FeC_{2}O_{4}$ in acid medium are oxidized by y moles of $KMnO_{4}$ . The ratio of x and y is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (A)

$Fe^{2 +} \rightarrow Fe^{3 +}+e$

$Mn^{7 +}+5e \rightarrow Mn^{2 +}$

$5moleFeSO_{4}\equiv 1moleMn^{7 +}$

$2moleFeSO_{4}\equiv \frac{2}{5}=x$

Also

$Fe^{2 +} \rightarrow Fe^{3 +}+e$

$\left(C^{3 +}\right)_{2} \rightarrow 2C^{4 +}+2e$

$Mn^{7 +}+5e \rightarrow Mn^{2 +}$

$5moleFeC_{2}O_{4}=3moleMn^{7 +}$

$2moleFeC_{2}O_{4}=\frac{3}{5}\times 2=y$

$\frac{x}{y}=\frac{1}{3}$