Multiple choice

$100: ml: 0.1 :M H_2O_2$ is mixed with $400: ml :0.5: M H_2O_2$. The volume of $0.1: M: KMnO_4$ (acidic) required to react with $H_2O_2$ mixture is :

  1. 840 ml

  2. 420 ml

  3. 630 ml

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total moles of H2O2 = (0.1 * 0.1) + (0.4 * 0.5) = 0.01 + 0.2 = 0.21 moles. Reaction: 2KMnO4 + 5H2O2 + 6H+ -> 2Mn2+ + 5O2 + 8H2O. Moles of KMnO4 = (2/5) * moles of H2O2 = 0.4 * 0.21 = 0.084 moles. Volume of 0.1 M KMnO4 = 0.084 / 0.1 = 0.84 litres = 840 ml.

AI explanation

Using the formula for the millimoles of a mixture, the total millimoles of H2O2 equals the sum of the products of volume and molarity for both solutions. We calculate 100 ml multiplied by 0.1 M plus 400 ml multiplied by 0.5 M, giving 10 millimoles plus 200 millimoles for a total of 210 millimoles. In an acidic medium, the redox reaction shows that 2 moles of KMnO4 react with 5 moles of H2O2, meaning the number of moles of KMnO4 needed is two-fifths of the moles of H2O2. Using the volume and molarity equality, 0.1 M multiplied by the required volume equals two-fifths of 210 millimoles, so the required volume is 42 divided by 0.05, which equals 840 ml.

Keep practicing — related questions