A 20.0 ,mL solution containing 0.2 ,g impure | vedclass

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(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.According to the l

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$85$

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(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 ightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.According to the law of equivalence,$Eq$ of $H_2O_2 = Eq$ of $KMnO_4$.Number of equivalents of $KMnO_4 = \frac{	ext{mass}}{	ext{equivalent weight}} = \frac{0.316}{158/5} = \frac{0.316 	imes 5}{158} = 0.01 \,Eq$.Since $Eq$ of $H_2O_2 = 0.01$,and the n-factor for $H_2O_2$ is $2$,the moles of $H_2O_2$ are $\frac{0.01}{2} = 0.005 \,mol$.Mass of pure $H_2O_2 = 0.005 	imes 34 = 0.17 \,g$.Purity of $H_2O_2 = \frac{	ext{mass of pure } H_2O_2}{	ext{mass of impure } H_2O_2} 	imes 100 = \frac{0.17}{0.2} 	imes 100 = 85\%$.

$A$ $20.0 \,mL$ solution containing $0.2 \,g$ impure $H_2O_2$ reacts completely with $0.316 \,g$ of $KMnO_4$ in acidic solution. The purity of $H_2O_2$ (in $\%$) is........... (mol. wt. of $H_2O_2 = 34$; mol. wt. of $KMnO_4 = 158$)

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