10 , mL of H_2O_2 solution (d = 1 , g/mL) whe | vedclass

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(B) The balanced chemical equation is: $H_2O_2 + 2H^+ + 2I^- ightarrow I_2 + 2H_2O$ From the stoichiometry,$1 \, 	ext{mole}$ of $H_2O_2$ produces $

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$0.669$

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(B) The balanced chemical equation is: $H_2O_2 + 2H^+ + 2I^- ightarrow I_2 + 2H_2O$ From the stoichiometry,$1 \, 	ext{mole}$ of $H_2O_2$ produces $1 \, 	ext{mole}$ of $I_2$. Molar mass of $H_2O_2 = 34 \, g/mol$. Molar mass of $I_2 = 254 \, g/mol$. Thus,$254 \, g$ of $I_2$ is produced by $34 \, g$ of $H_2O_2$. Therefore,$0.5 \, g$ of $I_2$ is produced by: $\frac{34}{254} 	imes 0.5 = 0.0669 \, g$ of $H_2O_2$. The mass of $10 \, mL$ of solution is $10 \, g$ (since $d = 1 \, g/mL$). Percentage purity of $H_2O_2 = \frac{	ext{mass of } H_2O_2}{	ext{mass of solution}} 	imes 100 = \frac{0.0669}{10} 	imes 100 = 0.669 \, \%$.

$10 \, mL$ of $H_2O_2$ solution $(d = 1 \, g/mL)$ when reacted with $KI$ solution produced $0.5 \, g$ of iodine. Calculate the percentage purity of $H_2O_2$ .............. $\%$

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