Photo AI

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

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⁺? 5 VO²⁺ + MnO₄⁻ + H₂O → 5 VO₃ + Mn²⁺ + 2 H⁺

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⁺? 5 VO²⁺ + MnO₄⁻ + H₂O → 5 VO₃ + Mn²⁺ + 2 H⁺ - AQA - A-Level Chemistry - Question 31 - 2020 - Paper 3

Step 1

96%

114 rated

Answer

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

[ \text{Moles of } MnO₄⁻ = \frac{0.01, \text{mol of VO}^2+}{5} = 0.002, \text{mol} ]

Step 2

99%

104 rated

Answer

We use the equation ( C = \frac{n}{V} ) to find the volume needed:

[ V = \frac{n}{C} ]

Substituting the values:

[ V = \frac{0.002, \text{mol}}{0.02, \text{mol dm}^{-3}} = 0.1, ext{dm}^3 ]

Since we need the volume in cm³:

[ V = 0.1, ext{dm}^3 \times 1000 = 100, ext{cm}^3 ]

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;