100 , mL of a given H_2O_2 sample in water (d | vedclass

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(D) The balanced chemical equation for the reaction between $H_2O_2$ and $KMnO_4$ in acidic medium is:$2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+}

Vedclass · Accepted Answer

Moles of $H_2O_2$ in the solution $= 5 	imes 10^{-4}$

Vedclass · Answer

(D) The balanced chemical equation for the reaction between $H_2O_2$ and $KMnO_4$ in acidic medium is:$2MnO_4^- + 5H_2O_2 + 6H^+ ightarrow 2Mn^{2+} + 5O_2 + 8H_2O$From the stoichiometry,$2 \, 	ext{moles of } KMnO_4$ react with $5 \, 	ext{moles of } H_2O_2$.Given: Volume of $KMnO_4 = 5 \, mL = 5 	imes 10^{-3} \, L$,Molarity of $KMnO_4 = 1/25 \, M = 0.04 \, M$.Moles of $KMnO_4 = 	ext{Molarity} 	imes 	ext{Volume} = 0.04 	imes 5 	imes 10^{-3} = 2 	imes 10^{-4} \, 	ext{moles}$.Using the stoichiometry,moles of $H_2O_2 = \frac{5}{2} 	imes (	ext{moles of } KMnO_4) = 2.5 	imes 2 	imes 10^{-4} = 5 	imes 10^{-4} \, 	ext{moles}$.Therefore,the correct statement is that the moles of $H_2O_2$ in the solution $= 5 	imes 10^{-4}$.

$100 \, mL$ of a given $H_2O_2$ sample in water (density $1.08 \, g/mL$) required $5 \, mL$ of $M/25 \, KMnO_4$ in acidic medium for its oxidation. Select the correct statement regarding $H_2O_2$ in the sample:

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