Question

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 solution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask. (Note: At the end point of the titration, the solution is a pale pink color.)

At a certain time during the titration, the rate of appearance of O2(g) was 1.0 x 10-3 mol/(L⋅s). What was the rate of disappearance of MnO4- at the same time. (please explain it)

Options
6.0 x 10-3 mol/(L⋅s)

A

4.0 x 10-3 mol/(L⋅s)

B

6.0 x 10-4 mol/(L⋅s)

C

4.0 x 10-4 mol/(L⋅s)

Answer 1

C. 4x10⁻⁴ mol / (Ls)

Step-by-step explanation:

Based in the reaction:

5 H₂O₂(aq) + 2 MnO₄⁻(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 O₂(g)

2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = 4x10⁻⁴ mol / (Ls)

Right answer is:

C. 4x10⁻⁴ mol / (Ls)