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Frankland Halide Reaction

In the reacti...

Question

In the reaction $N a_{2} S O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + H_{2} S O_{4} + 8 H C l$ the equivalent weight of $N a_{2} S O_{3}$ will be (M = molecular weight of $N a_{2} S O_{3}$ )

\frac{M}{4}

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\frac{M}{8}

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\frac{M}{1}

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\frac{W}{2}

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Solution

The correct option is B $\frac{M}{8}$
$N a_{2}_{2} O_{3} \to N a_{2} + 6 s O_{4}$
see the total change in oxidation number $= 4 \times 2 = 8$
$∴ E_{N a_{2} S_{2} O_{3}} = \frac{m o l . w t}{V . f} = \frac{M}{8}$

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Similar questions

N a_{2} S_{2} O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + 8 H C l

N a_{2} S_{2} O_{3}

=

N a_{2} S_{2} O_{3}

S_{2} O_{3}^{2 -} - ----- - + 5 H_{2} O + 4 {C l}_{2} ⟶ 2 S O_{4}^{2 -} + 8 {C l}^{-} + 10 H^{+}

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N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l

x

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N a_{2} S_{2} O_{3} + 5 H_{2} O + 4 C l_{2} ⟶ 2 N a H S O_{4} + 8 H C l

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K_{2} C r_{2} O_{7}

K_{2} C r_{2} O_{7}

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