Calculate the gram equivalent for the indicated reaction.

$H_{2} S ⟶ S$

17 g

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34 g

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51 g

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None of these

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Solution

The correct option is A $17 g$
$H_{2} S - 2 ⟶ S 0, n = 2.$

Molar mass of $H_{2} S$ = 34

$H e n c e, E q . w t = \frac{M o l a r m a s s}{2} = \frac{34}{2} = 17 g$

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

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25 g

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