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I was curious about this question too. But I looked at my lecture notes, and since [H2O] is in large excess, we consider the concentration to be unchanged and therefore, we don't include it in the equation.
Since water is in large excess, we assume its concentration to be relatively unchanged. That is, we aren't able to detect the change in water since it's probably something in the decimal units so it is essentially undetectable.
You leave out H2O because it serves as a solvent. Because of this, the concentration is in excess and you don't need to include it in the equilibrium constant expression. The same thing would happen if you had any other substance as a solvent, such as ethanol in some reactions
If you don't want to understand why: just remember we always leave out solids and liquids, molecules with the (s) and (l) subscript. If you want to know why, its because the reaction is happening in a solution of water, that means there are TREMENDOUS amounts of water compared to the other molecules so if any water ends up being used up or created in a reaction it really will make no difference. It's like adding adding or removing a droplet of water in a pool.
Like everyone has stated, it is because H20 is the solvent and thus in excess. This means that it will be on both sides of the reaction resulting in it canceling out during the calculations. However, water is not always left out only when it is the solvent.
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