A) The final volume is the addition of the volumes you used in the experiment To

Question

A) The final volume is the addition of the volumes you used in the experiment

Total volume = 12.5 + 12.5 + 5 + 25+ 10 = 65 ml or 0.065 L, divide it by 1000

b) You say all the S2O3 is consumed, so lets find how many moles you have

V = 10 ml or 0.01 L

Molarity = 0.01

moles available = Molarity * Volume = 0.01 * 0.01 = 0.0001 moles

For moles of I-, in the stoichiometry you have that for every 2 moles of S2O3 3 moles of I- are produced

moles of I- = 0.0001 * 3 / 2 = 0.00015 moles of I- produced

Molarity of I- produced = 0.00015 / 0.065 = 0.0023 M

These are my results the difference with you is that you took 1 mole of I- produced for every 2 moles of S2O3 and the reaction says that for every 2 moles of S2O3 you produce 3 moles of I-

Be carefull, I3- is triiodide on reactants side

I-  is iodide ion on product side

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Explanation / Answer

Can someone verify for me that I'm doing this correctly?

For data analysis, part b:

All S2O32- is consumed at the end of the reaction. Therefore, the moles of S2O32- consumed can be calculated using the equation below.
(moles S2O32-)consumed = Mstock x (Vstock)

My instructor states "The stock sodium thiosulfate (S2O3) is 0.01M

0.01M x 0.010 L = 0.0001moles S2O3"

So I did the rest of this accordingly (below), and it differs a lot from what's on this website. Any explanations would be appreciated, thanks!

Chemical Kinetics

Lab Results

Record the following lab data in the table below for trial 4. (done)

Data Analysis

Fill in the following table. Look below the table for instructions on how to calculate the values for each row of the table.

a) Calculate the final volume of the reaction mixture after the contents of beaker B are added to beaker A. Report your answer in liters.

b) All S2O32- is consumed at the end of the reaction. Therefore, the moles of S2O32- consumed can be calculated using the equation below.
(moles S2O32-)consumed = Mstock x (Vstock)

c) The I3- produced in reaction 1 reacts with S2O32- in reaction 2 as shown.
I3- (aq) 2S2O32- (aq) 3I- (aq) S4O62- (aq)

Therefore, for every mole of I3- produced, 2 moles of S2O32- have reacted.
(moles I3-)produced = (moles S2O32-)consumed / 2

d) [S2O32-]consumed = (moles S2O32-)consumed / (Vtotal)

e) Calculation is the same as d, but using I3- instead of S2O32-.

(a) volume of potassium iodide solution added to Erlenmeyer flask A in mL 12.5 mL (b) volume of potassium nitrate solution added to Erlenmeyer flask A in mL 12.5 mL (c) volume of starch solution added to Erlenmeyer flask A in mL 5 mL (d) volume of (NH4)2S2O8 solution added to Erlenmeyer flask B in mL 25 mL (e) volume of (NH4)2SO4 solution added to Erlenmeyer flask B in mL 0 mL (f) volume of Na2S2O3 solution added to Erlenmeyer flask B in mL 10 mL (g) the initial color of the solution after adding flask B to flask A grey/clear (h) time for color change 41 s (i) color of the final solution black

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