During electrolysis of water,the volume of O_ | vedclass

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(C) The balanced chemical equation for the electrolysis of water is: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$ According to the stoichiometry of the rea

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$4.48$

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(C) The balanced chemical equation for the electrolysis of water is: $2H_2O(l) ightarrow 2H_2(g) + O_2(g)$ According to the stoichiometry of the reaction,$2 	ext{ moles}$ of $H_2$ are produced for every $1 	ext{ mole}$ of $O_2$. Under the same conditions of temperature and pressure,the volume of gas is directly proportional to the number of moles. Therefore,the volume of $H_2$ liberated is twice the volume of $O_2$ liberated. $	ext{Volume of } H_2 = 2 	imes 	ext{Volume of } O_2 = 2 	imes 2.24 \ dm^3 = 4.48 \ dm^3$.

During electrolysis of water,the volume of $O_2$ liberated is $2.24 \ dm^3$. The volume of hydrogen liberated,under same conditions,will be .............. $dm^3$.

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