Question

2 moles of $FeS{O}_4$ in acid medium are oxidized by x moles of $KMn{O}_4$ , whereas 2 moles of $Fe{C}_2{O}_4$ in acid medium are oxidized by y moles of $KMn{O}_4$ . 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)

$F{e}^{2+}\to F{e}^{3+}+e$

$M{n}^{7+}+5e\to M{n}^{2+}$

$5moleFeS{O}_4\equiv 1moleM{n}^{7+}$

$2moleFeS{O}_4\equiv \frac{2}{5}=x$

Also

$F{e}^{2+}\to F{e}^{3+}+e$

${\left(C^{3+}\right)}_2\to 2C^{4+}+2e$

$M{n}^{7+}+5e\to M{n}^{2+}$

$5moleFe{C}_2{O}_4=3moleM{n}^{7+}$

$2moleFe{C}_2{O}_4=\frac{3}{5}\times 2=y$

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