I have to calculate the concentration of I2 produced (# of moles of I2 produced / L) based on the stoichiometric ratio in the following equation.

2S2O3^2- + I2 -> 2I^- + S4O6^2-

Chemistry(Urgent, please help) - DrBob222, Sunday, February 26, 2012 at 9:47pm
If you know the number of moles of S2O3^2- then the number of mols I2 will be 1/2 that. I suspect you have M and L of S2O3^- and M x L = moles S2O3^2-.

Chemistry(Urgent, please help) - DrBob222, Monday, February 27, 2012 at 12:47am
moles = M x L


M is the concentration correct?

Yes m is

Is L just the unit? Moles would be 3.33e-3M / L ?

Yes u remember where it says mxL this means that your multiplying m by l and l would be the unit

So to find the amount of I2 produced as I first posted, it would be 1/2 that of 3.33e-3 correct? So 0.0666? Sorry, its important that I understand this correct

Yes i tried your question and this is what i received

So the answer would be 6.66e-2 L*min? Is that the correct unit?

Sorry I actually got 0.00666 so it would be 6.6e-3 correct?

So I got 6.6e-3 L*min for the amount of I2 produced. Now I have to calculate the rate for each of the 5 flasks used, so is the answer going to be the same for all 5?? It says to calculate the rate of reaction for each run using # of moles of I2 produced/L*min as the unit.

Yes, in this context, M refers to the concentration of the solution in moles per liter (Molarity). So, if you have the concentration (M) and volume (L) of S2O3^2-, you can calculate the number of moles of S2O3^2- by multiplying the concentration (M) by the volume (L).

Once you have the number of moles of S2O3^2-, you can use the stoichiometric ratio in the balanced equation to determine the number of moles of I2 produced. The stoichiometric ratio in this case is 2 moles of S2O3^2- to 1 mole of I2. Therefore, the number of moles of I2 produced will be half of the number of moles of S2O3^2-.

To calculate the concentration of I2 produced, you need to divide the number of moles of I2 produced by the volume (L) of the solution.