[cisz]

A book I'm reading gives an example of a reaction between acetic acid and ethanol. It says that if 1 mole of each is used that there is .66 moles of ethyl acetate produced. Putting this into the equilibrium equation gives (.33 water)(.33 ethyl acetate)/(.17 ethanol)(.17 acetic acid) = 3.77.

They then say that by increasin the amount of ethanol to 2 moles that the amount of ethyl acetate can be calculated using the K_E  of 3.77.

Could someone explain how this calculation is done?

Thanks.

[Borek]

But you do know how to do equilibrium calculations in general? What an ICE table is?

[cisz]

No. I am new to this, but am learning.about it now.

I