What is the concentration of S2O82- remaining at 800 s is __________ M?

Question

The peroxydisulfate ion (S2O82-) reacts with the iodide ion in aqueous solution via the reaction: S2O8e (aq) + 3I --> 2SO4(aq) + I3(aq)

An aqueous solution containing 0.050 M of S2O82- ion and 0.072 M of I- is prepared, and the progress of the reaction followed by measuring [I-]. The data obtained is given in the table below.

Time(s)........ [I-] (M)
0..................0.072
400..............0.057
800..............0.046
1200............0.037
1600............0.029

What is the concentration of S2O82- remaining at 800 s is __________ M?

Answer 1

From the stoichiometrie of the reaction follows that for every three moles of iodide ions one mole of peroxydisulfate ions reacts away:
Δn(S₂O₈⁻) = n₀(S₂O₈⁻) - n(S₂O₈⁻) = 1/3 · Δn(I⁻)
The amount of peroxydisulfate ions remaining is:
n(S₂O₈⁻) = n₀(S₂O₈⁻) - 1/3 · Δn(I⁻) = n₀(S₂O₈⁻) - 1/3 ·(n₀(I⁻) - n(I⁻))
Since the total molar concentration can betaken constant for dilute aqueous solutions, you can rewrite in terms of concentrations:
[S₂O₈⁻] = [S₂O₈⁻]₀ - 1/3 ·( [I⁻]₀ - [I⁻] )

For t=800s you get:
[S₂O₈⁻] = 0.050M - 1/3·( 0.072M - 0.046M) = 0.0413M

Answer 2

I am not sure whether my method is right or not.
S2O8e (aq) + 3I --> 2SO4(aq) + I3(aq)
From the equation,
1 M of S2O8e react with 3 M of I produce 2 M SO4 and 1 M of I3.
0.024 M of S2O82- react with 0.072 M of I-.

After 800s, I- left 0.046 M.
This means 0.026 M of I- has reacted with 0.00867 M of S2O82-

Thus, the concentration of S2O82- remaining at 800s
is (0.050 - 0.00867) = 0.04133 M