What volume of O_{2(g)} measured at 1 atm an | vedclass

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(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 	o K_2SO_4 + 2MnSO_4 + 5O_2 + 8H_2O$. In a redox reaction,the n

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$0.56$

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(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 	o K_2SO_4 + 2MnSO_4 + 5O_2 + 8H_2O$. In a redox reaction,the number of equivalents of reactants are equal. Equivalents of $KMnO_4 = N 	imes V(L) = 0.5 	imes 0.1 = 0.05 \ eq$. Since equivalents of $KMnO_4$ = equivalents of $O_2$ produced,we have $0.05 \ eq$ of $O_2$. The n-factor for $O_2$ in this reaction (from $H_2O_2$ to $O_2$) is $2$. Moles of $O_2 = \frac{	ext{Equivalents}}{	ext{n-factor}} = \frac{0.05}{2} = 0.025 \ mol$. At $STP$ ($1 \ atm$ and $273 \ K$),$1 \ mol$ of gas occupies $22.4 \ L$. Volume of $O_2 = 0.025 	imes 22.4 = 0.56 \ L$.

What volume of $O_{2(g)}$ measured at $1 \ atm$ and $273 \ K$ will be formed by the action of $100 \ mL$ of $0.5 \ N \ KMnO_4$ on hydrogen peroxide in an acidic solution? The skeleton equation for the reaction is: $KMnO_4 + H_2SO_4 + H_2O_2 \to K_2SO_4 + MnSO_4 + O_2 + H_2O$.

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