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# Limiting reagent for an ester? watch

1. How do you know what the limiting reagent is?

E.g. for ethyl ethanoate is it ethanoic acid or ethanol?

10cm3 of ethanol and 12cm3 of ethanoic acid were used.

Is it just the one that has the least moles?
2. You are sort of correct, it would normally be the one with the least mol of reagent. BUT, in this case (and in most organic reactions), it is an equilibrium - the reaction is never going to go to completion. You'll end up doing a Kc calculation.

Anyhoo, you gave volumes, but no concentrations. Are they pure liquids? In which case, you'll need to know the values for the density of the two chemicals, so you can work out their masses and hence amounts...
3. (Original post by Pigster)
You are sort of correct, it would normally be the one with the least mol of reagent. BUT, in this case (and in most organic reactions), it is an equilibrium - the reaction is never going to go to completion. You'll end up doing a Kc calculation.

Anyhoo, you gave volumes, but no concentrations. Are they pure liquids? In which case, you'll need to know the values for the density of the two chemicals, so you can work out their masses and hence amounts...
Ah ok thanks

Yeah i have density and volume. So i do the relevant calculations and the one with the least moles is limiting?

Then i need to find % yield from the limiting reagent.

Is that moles of ester/moles of limiting reagent x 100 ?
4. Have you done Kc?

You can work out % yield, in terms of actual mass (or amount) of ester divided by the theoretical mass of ester that could be formed assuming it wasn't an equilibrium, but Kc is a much better way of expressing essentially the same thing.
5. (Original post by Pigster)
Have you done Kc?

You can work out % yield, in terms of actual mass (or amount) of ester divided by the theoretical mass of ester that could be formed assuming it wasn't an equilibrium, but Kc is a much better way of expressing essentially the same thing.
Yeah i have but how? (i know how to do it but how does it give the yield).

I was told to find actual masses etc. - so i think it would be better to do the other method ohh yeah i forgot oops (actual/theoretical)

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