The volume (in mL) of 10 volume H_2O_2 soluti | vedclass

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Question

(C) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO

Vedclass · Accepted Answer

$224$

Vedclass · Answer

(C) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 ightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$. From the stoichiometry,$2 	ext{ moles of } KMnO_4$ react with $5 	ext{ moles of } H_2O_2$. Moles of $KMnO_4 = 	ext{Molarity} 	imes 	ext{Volume (L)} = 0.4 	imes 0.2 = 0.08 	ext{ mol}$. Moles of $H_2O_2$ required = $\frac{5}{2} 	imes 0.08 = 0.2 	ext{ mol}$. Mass of $H_2O_2$ = $0.2 	imes 34 = 6.8 	ext{ g}$. $10$ volume $H_2O_2$ means $1 	ext{ mL}$ of solution gives $10 	ext{ mL}$ of $O_2$ at $STP$. The strength of $10$ volume $H_2O_2$ is $\frac{68}{22.4} 	imes 10 \approx 30.36 	ext{ g/L}$ or $3.036 \%$. Using the relation: $	ext{Volume strength} = 5.6 	imes 	ext{Normality}$. $10 = 5.6 	imes N \Rightarrow N = \frac{10}{5.6} = 1.786 	ext{ N}$. Since $n$-factor for $H_2O_2$ is $2$,Molarity $M = \frac{1.786}{2} = 0.893 	ext{ M}$. Equating milliequivalents: $N_1V_1 = N_2V_2$. $N_{KMnO_4} = 0.4 	imes 5 = 2 	ext{ N}$. $2 	imes 200 = 1.786 	imes V_{H_2O_2} \Rightarrow V_{H_2O_2} = \frac{400}{1.786} \approx 224 	ext{ mL}$.

The volume (in $mL$) of $10$ volume $H_2O_2$ solution required to completely react with $200 \ mL$ of $0.4 \ M \ KMnO_4$ solution in acidic medium is

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