The volume of oxygen liberated from 0.68 g o | vedclass

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(B) The balanced chemical equation for the decomposition of $H_2O_2$ is:$2H_2O_2 ightarrow 2H_2O + O_2$From the stoichiometry:$2 	imes 34 \ g$ of $

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(B) The balanced chemical equation for the decomposition of $H_2O_2$ is:$2H_2O_2 ightarrow 2H_2O + O_2$From the stoichiometry:$2 	imes 34 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$ at $STP$.$68 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$.Therefore,the volume of $O_2$ produced from $0.68 \ g$ of $H_2O_2$ is:$	ext{Volume} = \frac{22400 \ mL 	imes 0.68 \ g}{68 \ g} = 224 \ mL$.

The volume of oxygen liberated from $0.68 \ g$ of $H_2O_2$ is ............. $mL$ at $STP$.

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