Calculate the gram equivalent for the given reaction.

$H_{2} O_{2} ⟶ H_{2} O$

8.5 g

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17 g

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34 g

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None of these

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Solution

The correct option is C $17 g$
$H_{2} O_{2} - 1 ⟶ {2 H}_{2} O - 2,$

Hence, $E q . w t . = 34 / 2 = 17 g$ (O.N of two $O -$ atoms have been considered).

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Similar questions

H_{2} O_{2} (⟶ O_{2})

H_{2} S ⟶ S

H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

H_{2} O_{2} \to H_{2} O + O_{2}

S O_{2}

S O_{2} + H_{2} S \to S + H_{2} O

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