The equivalent mass of $M n_{3} O_{4}$ in $M n_{3} O_{4} + 2 F e S O_{4} + 4 H_{2} S O_{4} ⟶ F e_{2} {(S O_{4})}_{3} + 3 M n S O_{4} + H_{2} O$ is $\frac{M}{x}$ . Then, the value of $x$ is :

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Solution

$M n_{3} O_{4} + 2 F e S O_{4} + 4 H_{2} S O_{4} ⟶ F e_{2} {(S O_{4})}_{3} + 3 M n S O_{4} + H_{2} O$
+8/3 +2
As we know,
$E q u i v a l e n t w e i g h t = \frac{M o l a r w e i g h t}{n - f a c t o r} = \frac{M}{2}$
Since, change in oxidation state = $\frac{2}{3}$ . For 3 Mn atoms, change = $2$ .

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F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

F e C_{2} O_{4} ⟶ F e^{3 +} + C O_{2}

F e C_{2} O_{4}

\frac{M}{x}

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A s_{2} O_{3}

A s_{2}^{3 +} ⟶ A s_{2}^{5 +}

\frac{M}{x}

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M n_{3} - --- - O_{4}

A s O_{2}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ B r^{-} + A s O_{4}^{-} + H_{2} O

B r O_{3}^{-} = \frac{M}{x}

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