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What is the minimum volume, in cm³, of 0.02 mol dm⁻³ KMnO₄ solution needed to oxidise 0.01 mol of VO₂⁺? 5VO₂⁺ + MnO₄⁻ + H₂O → 5VO₃⁻ + Mn²⁺ + 2H⁺ - AQA - A-Level Chemistry - Question 31 - 2020 - Paper 3

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Question 31

What-is-the-minimum-volume,-in-cm³,-of-0.02-mol-dm⁻³-KMnO₄-solution-needed-to-oxidise-0.01-mol-of-VO₂⁺?--5VO₂⁺-+-MnO₄⁻-+-H₂O-→-5VO₃⁻-+-Mn²⁺-+-2H⁺-AQA-A-Level Chemistry-Question 31-2020-Paper 3.png

What is the minimum volume, in cm³, of 0.02 mol dm⁻³ KMnO₄ solution needed to oxidise 0.01 mol of VO₂⁺? 5VO₂⁺ + MnO₄⁻ + H₂O → 5VO₃⁻ + Mn²⁺ + 2H⁺

Worked Solution & Example Answer:What is the minimum volume, in cm³, of 0.02 mol dm⁻³ KMnO₄ solution needed to oxidise 0.01 mol of VO₂⁺? 5VO₂⁺ + MnO₄⁻ + H₂O → 5VO₃⁻ + Mn²⁺ + 2H⁺ - AQA - A-Level Chemistry - Question 31 - 2020 - Paper 3

Step 1

Calculate Moles of KMnO₄ Required

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Answer

From the balanced equation, 5 moles of VO₂⁺ react with 1 mole of MnO₄⁻. Therefore, for 0.01 moles of VO₂⁺, the moles of MnO₄⁻ required is:

nMnO4−=0.01 mol VO₂⁺5=0.002 mol MnO₄⁻n_{MnO₄^-} = \frac{0.01 \text{ mol VO₂⁺}}{5} = 0.002 \text{ mol MnO₄⁻}

Step 2

Calculate Volume of KMnO₄ Solution Needed

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Answer

Using the concentration formula, where concentration (C) = moles (n) / volume (V), we can rearrange to find volume:

V=nC=0.002 mol0.02 mol dm−3=0.1 dm3V = \frac{n}{C} = \frac{0.002 \text{ mol}}{0.02 \text{ mol dm}^{-3}} = 0.1 \text{ dm}^3

To convert to cm³:

0.1 dm3=100 cm30.1 \text{ dm}^3 = 100 \text{ cm}^3

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