03 + 3I- + H20 --> O2 + I3^- + 2OH-
I3^- + 2S2O3^2- ==> S4O6^2- + 3I^-
mols S2O3^2- = N x L = ? which you have done correctly.
1/2 moles S2O3^2- = moles I3^- = mols O3
moles O3 x molar mass = grams O3 and convert that to mg.
I believe that gets the 35.5 mg.
03 + 3I- + H20 --> O2 + I3^- + 2OH-
A 1.00-L bulb of air containing O3 produced by an electric spark was treated with 25mL of 2M KI, shaken well, and left closed for 30 minutes so that all O3 would react. The aqueous solution was then drained from the bulb, acidified with 2mL of 1 M H2SO4 and required 29.33mL of 0.05044 M S2O3^2- for titration of I3^-
Calculate the mass of O3 in the 1.00-L bulb.
I started by multiplying 0.02933L * 0.05044 M S2O3 to get moles...but then I kind of got stuck. Not sure if this is right or what's the next step.
The answer is 35.5mg
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