The decomposition of H_2O_2 produces O_2. If | vedclass

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(D) The balanced chemical equation for the decomposition of hydrogen peroxide is: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$ First,calculate the mole

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(D) The balanced chemical equation for the decomposition of hydrogen peroxide is: $2H_2O_2(aq) ightarrow 2H_2O(l) + O_2(g)$ First,calculate the moles of $O_2$ produced: $Moles \ of \ O_2 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{48 \ g}{32 \ g \ mol^{-1}} = 1.5 \ mol$ Since $1 \ minute$ has passed,the rate of formation of $O_2$ is $1.5 \ mol \ min^{-1}$. From the stoichiometry of the reaction,$1 \ mol$ of $O_2$ is produced along with $2 \ mol$ of $H_2O$. Therefore,the rate of formation of $H_2O = 2 	imes (	ext{rate of formation of } O_2) = 2 	imes 1.5 \ mol \ min^{-1} = 3.0 \ mol \ min^{-1}$.

The decomposition of $H_2O_2$ produces $O_2$. If at a certain time,$48 \ g$ of $O_2$ is produced in $1 \ minute$,then the rate of formation of water at that time will be .......... $mol \ min^{-1}$.

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