Chemistry How Many Moles Were Consumed in a Titration with 5.29 ml of 0.256 M Na2S2O3?

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In the titration discussion, participants analyze the calculation of moles consumed from a titration involving 5.29 ml of 0.256 M Na2S2O3. The correct approach involves using the formula M = moles/volume to find the moles of S2O3 consumed, which is calculated as 0.256 mol/L multiplied by 0.00529 L, resulting in 0.00135 moles. The confusion arises from the stoichiometry of the reaction, where participants mistakenly multiply by two when calculating moles of S2O3 due to its presence in the balanced equation. The final consensus indicates that the correct moles of OCl in the sample should be derived from the stoichiometric ratios rather than direct multiplication. Accurate understanding of concentration, volume, and stoichiometry is essential for solving titration problems effectively.
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Homework Statement



The overall chemical equation for the titration reaction is:
OCl(aq) + 2 S2O3(aq) + 2H+(aq) 6 Cl(aq) + S2O6(aq) + H2O(R)

If a titration requires 5.29 ml 0.256 M Na2S2O3,
(i) how many moles of S2O32&(aq) were consumed in the titration, and
(ii) how many moles of OCl&(aq) were in the sample?


Homework Equations





The Attempt at a Solution



Ok so what i did was take .245/52.9 and multiplied it by 2 to get .0096 moles of S2O3 consumed...is that right?
 
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No.

Try to explain what you did and why.

And using numbers that were given in the question won't hurt.
 
I just realized there is supposed to be an arrow in between the 2H and the Cl; that is where the equation splits into reactants and products.

I'm looking over it again and trying something different. I messed up in the math the first go around

.256/.00529 L = 48.4 mols X 2 (because there are 2 mols S2O3) and that gives me 96.8 mols S2O3; part i

Shouldn't part (ii) just be the 48.4 mols from the previous?
 
You are still wrong. Check your units.
 
M = mols/volume

I know M and the volume, so it should be M * volume..not divided..right?

.256 * .00529 = .00135

.00135 * 2 = .0027 mols. Is that right?
 
Better, but still wrong. You got concentration&volumes&moles part right. Why do you multiply by two if you are calculating number of moles of S2O32-?
 
I multiplied by 2 because there are 2 S2O3 in the equation. Doesn't that mean you multiply it by two so that your ratio is correct?
 
Think about it - you put a mole of substance into the baker, but as reaction equation have a 2 in the equation, that means you put 2 moles in the baker? That's what you did now.
 
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