A sample of $F e S O_{4}$ and $F e C_{2} O_{4}$ is dissolved in $H_{2} S O_{4} .$ The complete oxidation of sample required 8/3 equivalents of ${KMnO}_{4} .$ After this the reaction mixture was reduced by zinc. On again oxidation by ${KMnO}_{4}$ required 5/3 equivalents. The mole ratio of $F e S O_{4}$ and $F e C_{2} O_{4}$ is

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answer is B.

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Detailed Solution

Let moles of $F e S O_{4} = x$
Moles of $F e C_{2} O_{4} = y$

$\begin{array}{l} x + 3 y = \frac{8}{3} \\ x + y = \frac{5}{3} \\ x = \frac{7}{6}; y = \frac{1}{2} \\ \frac{x}{y} = \frac{7}{3} \end{array}$

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