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# Calculate moles

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1. Hi everyone,

We have just done a tritation calculation for an equilibrium reaction. This involved an acid catalyst, mixed with ethanol and ethanoic acid.

I know the number of moles of the catalyst, and the number of moles of NaOH which reacted with the catalyst and any remaining ethanoic acid, but how I calculate tye number of moles of ethanoic acid which reacted with the alcohol?

Mol. Of HCl = 0.015
Mol. Of NaOH = 0.003725
2. (Original post by Electrogeek)
Hi everyone,

We have just done a tritation calculation for an equilibrium reaction. This involved an acid catalyst, mixed with ethanol and ethanoic acid.

I know the number of moles of the catalyst, and the number of moles of NaOH which reacted with the catalyst and any remaining ethanoic acid, but how I calculate the number of moles of ethanoic acid which reacted with the alcohol?

Mol. Of HCl = 0.015
Mol. Of NaOH = 0.0003725
write out a balanced equation which will tell you how many moles of what reacted with what
3. (Original post by thefatone)
write out a balanced equation which will tell you how many moles of what reacted with what
Is the equation for the HCl or for the ethanoic acid?
4. (Original post by Electrogeek)
Is the equation for the HCl or for the ethanoic acid?
this looks like a classic acid + alcohol / ester + water equilibrium calculation

If you know how much acid catalyst you had (I assume from what you have written above that you used concentrated HCl) and you know how much NaOH you needed to titrate ALL acid present then

moles of ethanoic acid (that didn't react with the alcohol) = moles of NaOH - moles of HCl

EDIT: Your given experimental values don't make any sense. Even if ALL of the ethanoic acid had reacted you would need the same number of moles of NaOH as HCl.
5. (Original post by TeachChemistry)
this looks like a classic acid + alcohol / ester + water equilibrium calculation

If you know how much acid catalyst you had (I assume from what you have written above that you used concentrated HCl) and you know how much NaOH you needed to titrate ALL acid present then

moles of ethanoic acid (that didn't react with the alcohol) = moles of NaOH - moles of HCl

EDIT: Your given experimental values don't make any sense. Even if ALL of the ethanoic acid had reacted you would need the same number of moles of NaOH as HCl.
I thought that was what I had to do - and thats why I became confused because it would give a negative value. Thanks. :-)
6. (Original post by Electrogeek)
I thought that was what I had to do - and thats why I became confused because it would give a negative value. Thanks. :-)
What was the concentration and titre for your NaOH solution?
7. (Original post by TeachChemistry)
What was the concentration and titre for your NaOH solution?
0.5 mol NaOH, with a titre of 7.45 cm^3
8. (Original post by Electrogeek)
0.5 mol NaOH, with a titre of 7.45 cm^3
And looking at your figures you used 1.25 cm3 of conc. HCl?
9. (Original post by TeachChemistry)
And looking at your figures you used 1.25 cm3 of conc. HCl?
We found the problem - the molarity of NaOH wasn't for the volume if our solution - we had to times the value by 5, and then it all works out. Turns out when our teacher produced the sheet they accidently mixed up the values with another question (which we compared our results to)

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