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Stoichiometric Calculations

Two moles of ...

Question

Two moles of $F e S O_{4}$ are oxidized by $x$ moles of $K M n O_{4}$ in acid medium, whereas $3$ moles of $F e C_{2} O_{4}$ are oxidized by $y$ moles of $K_{2} C r_{2} O_{7}$ in acid medium. The value of $\frac{x}{y}$ is :

\frac{6}{5}

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\frac{4}{15}

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\frac{18}{15}

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\frac{3}{5}

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Solution

The correct option is D $\frac{4}{15}$
Reaction:
$2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O$
and,
$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 2 F e C_{2} O_{4} \to F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 C O_{2} + 7 H_{2} O$
So here, $\frac{x}{y} = \frac{0.4}{1.5} = \frac{4}{15}$

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