[cvc121]

Hi,

Given the following information and equation, how many moles of S2O8^2- are not used if all of the I^- is completely changed to I2?

2I^- + S2O8^2- I2 + 2SO4^2-

Not sure if all of this information is necessary but...you have 0.01L of 0.20M KI solution (0.002 moles of I^-) and 0.02L of 0.10M (NH4)2S2O8 solution (0.002 moles of S2O8^2-).

What I did was use the molar ratio to find the number of moles of S2O8^2- given that I have 0.002 moles of I^-

Moles of S2O8^2- = 0.002 mol/2 = 0.001 mol of S2O8^2- not used

I am not too sure if what I solved is what the question is asking for. Can anyone clarify?
Thanks! All help is very much appreciated.

[thetada]

Looks good. The same number of moles of each are mixed together but you only need half as many moles of S2O82- so 0.001 moles will be left over.