$H_{2} O_{2} (⟶ O_{2})$
The grams of one equivalent each for the indicated reaction.

17 g

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51 g

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34 g

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None of these

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Solution

The correct option is A $17 g$
$H_{2} O_{2} - 2 ⟶ O_{2} 0, n f a c t o r = 2.$
$H e n c e, E q . w t . = 17 g$ (O.N of two $O -$ atoms have been considered)

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Similar questions

H_{2} O_{2} ⟶ H_{2} O

H_{2} S ⟶ S

H_{2}

O_{2}

If 2 moles of H₂ and O₂ are available for the reaction, identify the limiting reagent for the reaction,

2H₂ + O₂ → 2H₂O

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

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