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For the reaction, a 0.9621 g sample of an unknown aqueous hydrogen peroxide solution required 11.07 mL of 0.0200 M KMnO4. Calculate the mass percent peroxide (to two decimal places) in the unknown sample. The molar mass of H_{2}O_{2}

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A 0.9621 g sample, whose mass percent of hydrogen peroxide is 1.17% requires 11.07 mL of 0.0200 M KMnO₄ to react completely.

Let's consider the balanced equation for the reaction between H₂O₂ and KMnO₄.

2 KMnO₄ + 3 H₂O₂ → 3 O₂ + 2 MnO₂ + 2 KOH + 2 H₂O

11.07 mL of 0.0200 M KMnO₄ react. We can calculate the reacting mass of H₂O₂ considering that:

  • The molar ratio of KMnO₄ to H₂O₂ is 2:3.
  • The molar mass of H₂O₂ is 34.01 g/mol.

[tex]0.01107L\ KMnO_4 \times \frac{0.0200mol KMnO_4}{1L\ KMnO_4} \times \frac{3molH_2O_2}{2molKMnO_4} \times \frac{34.01gH_2O_2}{1molH_2O_2} =0.0113gH_2O_2[/tex]

0.0113 g of H₂O₂ are in 0.9621 g of the sample. The mass percent of H₂O₂ in the sample is:

[tex]\% H_2O_2 = \frac{0.0113g}{0.9621g} \times 100\% = 1.17\%[/tex]

A 0.9621 g sample, whose mass percent of hydrogen peroxide is 1.17% requires 11.07 mL of 0.0200 M KMnO₄ to react completely.

Learn more: brainly.com/question/9743981

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