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

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.

To calculate the mass of O3 in the 1.00-L bulb, we need to use the given information and perform a series of calculations. Here is a step-by-step approach:

1. Determine the number of moles of S2O3^2- used for titration:
- Volume of S2O3^2- solution used for titration: 29.33 mL = 0.02933 L
- Concentration of S2O3^2- solution: 0.05044 M
- Moles of S2O3^2- used = Volume × Concentration = 0.02933 L × 0.05044 M

2. Calculate the number of moles of I3^- reacted:
- According to the balanced equation, 1 mole of I3^- reacts with 1 mole of S2O3^2-
- Moles of I3^- = Moles of S2O3^2- used (from step 1)

3. Convert moles of I3^- to moles of O3:
- According to the balanced equation, 1 mole of O3 reacts with 1 mole of I3^-
- Moles of O3 = Moles of I3^- (from step 2)

4. Calculate the molar mass of O3:
- The molar mass of O3 is 48.00 g/mol

5. Determine the mass of O3:
- Mass of O3 = Moles of O3 × Molar mass of O3 = Moles of O3 × 48.00 g/mol

Once you have completed these calculations, you will find that the mass of O3 in the 1.00-L bulb is 35.5 mg.