What is the volume (in mL) of 20 vol H_2O_2 r | vedclass

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Question

(A) The balanced chemical equation for the reaction between $H_2O_2$ and acidified $KMnO_4$ is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The balanced chemical equation for the reaction between $H_2O_2$ and acidified $KMnO_4$ is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$ From the stoichiometry,$2 	ext{ moles of } KMnO_4$ react with $5 	ext{ moles of } H_2O_2$. Moles of $KMnO_4 = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 0.02 	imes 0.5 = 0.01 	ext{ mol}$. Moles of $H_2O_2$ required $= \frac{5}{2} 	imes 0.01 = 0.025 	ext{ mol}$. Normality of $20 	ext{ vol } H_2O_2 = \frac{20}{5.6} \approx 3.57 	ext{ N}$. Since the $n$-factor for $H_2O_2$ in this reaction is $2$,Molarity of $H_2O_2 = \frac{3.57}{2} = 1.785 	ext{ M}$. Volume of $H_2O_2 = \frac{	ext{Moles}}{	ext{Molarity}} = \frac{0.025}{1.785} \approx 0.014 	ext{ L} = 14 	ext{ mL}$. Alternatively,using the law of equivalence: $N_1V_1 = N_2V_2$. Normality of $KMnO_4 = 0.02 	imes 5 = 0.1 	ext{ N}$. $3.57 	imes V_1 = 0.1 	imes 500 \implies V_1 = \frac{50}{3.57} \approx 14 	ext{ mL}$. Thus,option $(A)$ is the correct answer.

What is the volume (in $mL$) of $20$ vol $H_2O_2$ required to completely react with $500 \ mL$ of $0.02 \ M$ acidified $KMnO_4$ solution?

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