Ok, perhaps a more challenging Chemistry question... this one has stumped me for 4 days now.?

The SO2 present in the air is mainly responsible for the phenomenon of acid rain. The concentration of SO2 is determined by titrating against a standard permanganate solution in acid with the following unbalanced redox reaction:

SO2 + MnO4- -- SO4^2- + MN^2+

Calculate the number of grams of SO2 in a sample of air if 7.37 mL of 0.00800M KMNO4 solution are required for the titration.

Any help is appreciated

Answer:
Oxidation: SO2 --> SO4^2-
Reduction: MnO4^- --> Mn^2+

2 H2O + SO2 --> SO4^2- + 4 H^+ + 2 e^-
5 e^- + 8H^+ MnO4^- --> Mn^2+ + 4 H2O

Multiply top reaction by 5 and bottom reaction by 2

10 H2O + 5 SO2 --> 5 SO4^2- + 20 H^+ + 10 e^-
10 e^- + 16 H^+ + 2 MnO4^- --> 2 Mn^2+ + 8 H2O

Add and cancel electrons, water and H^+

2 H2O + 5 SO2 + 2 MnO4^- --> 5 SO4^2- + 4 H^+ + 2 Mn^2+

So, 5 moles SO2 corresponds to 2 moles of MnO4^-

0.00800 M X 0.00737 L = 0.0000590 moles MnO4^-

0.0000590 moles MnO4^- X (5 moles SO2/2 moles MnO4^-) = 0.000147 moles SO2

0.000147 moles SO2 X (64.0 g/mole SO2) = 0.00943 g SO2
Here is a much easier way to do this:

0.00800Moles/LKMnO4x0.00737L=5...

Considering there is a 1to1 ratio you have 5.90x10-5moles of SO2. Now convert moles to grams.

5.90x10-5molesSO2x(64.0098gSO2...

3.78x10-3 grams SO2
I got the same answer as the second guy and realized this needed to be balanced. Trust in the professor... hes got it right. see for yourself be applying the 5:2 ratio to the second answer here and its right.

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