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Part B Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)→3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 17.8 mL of the KMnO4 solution? Express your answer with the appropriate units. View Available Hint(s) mass of H2O2H 2 O 2 = Previous Answers Incorrect; Try Again; 5 attempts remaining Provide Feedback Incorrect. Incorrect; Try Again; 5 attempts remaining. No additional feedback.

User Clarity
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Answer:

0,508g of H₂O₂

Step-by-step explanation:

For the reaction:

2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)

2 moles of KMnO₄ react with 1 mol of H₂O₂.

In the titration, moles of KMnO₄ required were:

1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:

0,0299 moles of KMnO₄×
(1molH_(2)O_(2))/(2molKMnO_(4)) = 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:

0,01495 moles of H₂O₂×
(34,01g)/(1molH_(2)O_(2)) = 0,508g of H₂O₂

User Sabith
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