Final answer:
The mole ratio of H2O2 to MnO4− is 5:2. Using this ratio, we can determine that 0.0045 moles of potassium permanganate will oxidize approximately 0.01125 moles of hydrogen peroxide. The option closest to this value is not listed, but the calculated amount should be used for accurate stoichiometric calculations.
Step-by-step explanation:
The mole ratio of hydrogen peroxide (H2O2) to permanganate ion (MnO4−) in the balanced chemical equation of their redox reaction in an acidic solution is key to answering the question. The typical balanced equation is:
5 H2O2(aq) + 2 MnO4−(aq) + 6 H+(aq) → 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)
Using the provided balanced equation, the mole ratio of H2O2 to MnO4− is 5:2. This means that it takes 5 moles of H2O2 to react with 2 moles of MnO4−. To find the amount of hydrogen peroxide oxidized by 0.0045 moles of potassium permanganate (which yields MnO4− in solution), set up the ratio:
(5 mol H2O2) / (2 mol MnO4−) = (x mol H2O2) / (0.0045 mol MnO4−)
Solving for x gives you 0.01125 mol of H2O2, which is not an option in the multiple-choice answers. However, looking at the provided options and considering significant figures, the answer can be approximated to 0.01125 moles, which in the context of significant figures can be rounded to 0.011 moles if that option were present.