What is the approximate volume (in mL) of 10 | vedclass

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(A) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is: $2 \ KMnO_4 + 3 \ H_2SO_4 + 5 \ H_2O_2 \long

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$56.05$

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(A) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is: $2 \ KMnO_4 + 3 \ H_2SO_4 + 5 \ H_2O_2 \longrightarrow K_2SO_4 + 2 \ MnSO_4 + 8 \ H_2O + 5 \ O_2$ From the stoichiometry,$2 \ 	ext{moles of } KMnO_4$ react with $5 \ 	ext{moles of } H_2O_2$. Given: $n(KMnO_4) = 	ext{Molarity} 	imes 	ext{Volume} = 0.02 \ M 	imes 1 \ L = 0.02 \ 	ext{moles}$. Therefore,$n(H_2O_2) = \frac{5}{2} 	imes 0.02 = 0.05 \ 	ext{moles}$. Mass of $H_2O_2 = 0.05 \ 	ext{mol} 	imes 34 \ g/mol = 1.7 \ g$. $10 \ 	ext{vol } H_2O_2$ means $1 \ L$ of $H_2O_2$ solution gives $10 \ L$ of $O_2$ at $STP$. Since $1 \ L$ of $H_2O_2$ contains $30.3 \ g$ of $H_2O_2$ (as $68 \ g$ of $H_2O_2$ gives $22.4 \ L$ of $O_2$,so $10 \ L$ of $O_2$ corresponds to $\frac{68 	imes 10}{22.4} \approx 30.35 \ g$),the concentration is $\frac{30.35 \ g}{1000 \ mL}$. Volume of $H_2O_2$ solution required $= \frac{1.7 \ g}{30.35 \ g/L} \approx 0.056 \ L = 56 \ mL$.

What is the approximate volume (in $mL$) of $10$ vol $H_2O_2$ solution that will react completely with $1 \ L$ of $0.02 \ M \ KMnO_4$ solution in acidic medium?

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