The Student Room Group
Reply 1
You need the ratios of products and reactants, but in this case it looks like 1 mole + 1 mole -> 1 mole + 1 mole.

In that case, you can see that you have used up 2.5 moles of methanoic acid - this will have reacted with 2.5 moles of ethanol to give 2.5 moles of each product. Therefore, as 2.5 moles of ethanol are used up, there will be 6.25-2.5 moles in the mixture - i.e. 3.75 moles.
Reply 2
Its a reaction of an Organic acid and a primary Alcohol


C2H5OH(aq) + HCOOH(aq) <----> HCOOC2H5(aq) + H2O(l)

Kc= [HCOOC2H5]/[C2H5OH][HCOOH]
Reply 3
AVAINTEX
Its a reaction of an Organic acid and a primary Alcohol


C2H5OH(aq) + HCOOH(aq) <----> HCOOC2H5(aq) + H2O(l)

Kc= [HCOOC2H5]/[C2H5OH][HCOOH]


Since its a weak acid.it will not ionize completely. We will assume that the volume is 1L

and it is all yours.
Reply 4
nigel_s
You need the ratios of products and reactants, but in this case it looks like 1 mole + 1 mole -> 1 mole + 1 mole.

In that case, you can see that you have used up 2.5 moles of methanoic acid - this will have reacted with 2.5 moles of ethanol to give 2.5 moles of each product. Therefore, as 2.5 moles of ethanol are used up, there will be 6.25-2.5 moles in the mixture - i.e. 3.75 moles.


Thanks.

Usually when I work these out myself I set it out like so:

EQUATION:

Initial Conc:

Equilibria Conc:

Reacted Conc:

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