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Frankland Halide Reaction

In the reacti...

Question

In the reaction, $V O + F e_{2} O_{3} ⟶ F e O + V_{2} O_{5}$ , the equivalent mass of $V_{2} O_{5}$ is equal to its :

m o l a r m a s s

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\frac{m o l a r m a s s}{8}

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\frac{m o l a r m a s s}{6}

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none of these

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Solution

The correct option is C $\frac{m o l a r m a s s}{6}$
$V O + F e_{2} O_{3} ⟶ F e O + V_{2} O_{5}$
+2 +3 +2 +5
Change in oxidation state of $V = 5 - 2 = 3$ .
So, equivalent weight of $V_{2} O_{5} = \frac{m o l a r m a s s}{6}$

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Similar questions

V O + F e_{2} O_{3} \to F e O + V_{2} O_{5}

V_{2} O_{5}

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

V_{2} O_{5}

5.75

F e_{2} O_{3}

V O + F e_{2} O_{3} \to F e O + V_{2} O_{5}

= 56

= 51

= 16

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

\frac{M}{3}

= \frac{Molar mass}{n-factor}

{C r_{2} O_{7}}^{2 -} \to C r^{3 +}

{C r_{2} O_{7}}^{2 -} = \frac{Molar Mass}{3}

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