Okay, so I was browsing through an AP chemistry book because I really want to major in either chemistry of physics and want to take advantage over thanksgiving break to get ahead on chemistry because I really love it ^^. So I was wondering if someone could check my work on this problem. Thanks ^^!! I. 2(Mn2)+ + 4(OH-) + (O2) -> 2(MnO2) + 2(H2O) II. (MnO2) + 2(I-) + 4(H+) -> Mn(2+) + (I2) + 2(H2O) III. 2((S2O3)2-)) + (I2) -> ((S4O6)2-)) + 2(I-) The amount of oxygen dissolved in water can be determined by titration. First, (MnSO4) and NaOH are added to a sample of water to convert all of the dissolved oxygen to (MnO2), as shown in equation I above. Then (H2SO4) and KI are added and the reaction represented by equation II proceeds. Finally, the (I2) that is formed is titrated with standard sodium thiosulfate (Na2S2O3), according to equation III. 1) A student found that a 50 mL sample of water required 4.86 mL of .0112 M (Na2S2O3) to reach the equivalence point. Calculate the number of moles of oxygen dissolved in this sample. After using Hess' Law, I got 4(OH-) + (O2) + 8(H+) + 4((S2O3)2-)) -> 6(H2O) + 2((S4O6)2-) Anyway.. I did this 4.86 mL x (11.2 mol/mL) = 54.4 mol (Na2S2O3) Then I converted that to moles of oxygen by dividing by four (what I got from Hess' law) So 13.6 moles (O2) Then I divided by .05L, the amount of water we are given. Resulting in 272.2 mol (O2/L) My question here is, did I do the set up correctly? There were more L/mL units than I'm used to, so I got confused with how to cancel them or set them up in equations. If not, any advice would be very helpful and appreciated ^^. 2) How would the results above be affected if some (I2) were lost before the ((S2O3)2-) was added? Explain. I said the results would yield more moles of oxygen because you would need more of everything else to compensate for the lost. Though, I'm not really sure because when I did Hess' law iodine was gone and canceled out. 3) What volume of dry (O2) measured at 25*C and 1 atm of pressure would have to be dissolved in 1 L of pure water in order to prepare a solution of the same concentration that was obtained in (1)? Well, if I did 1 wrong, then I'm going to do this wrong.. but here's my dilemma. I am continually confused with all the liter information I keep getting.. But here I go.. The density of water is 1 g/L, and you have 1 L of water, therefore you have 1 g of water or 1/18 moles of water. (1/18) x (1 mol 02 / 6mol H20) = 1/108 mol 02 Now.. the mol's of oxygen in part I were quite large.. and this is very small.. leading me to believe I made some huge error. So I'm requesting any guidance if anyone is willing to help. Thanks again ^^.. even just for reading this!