Understanding current, voltage and resistance
But what happens if we do not add the resistor? ... The LED drops 2.5V, but the battery's potential is 9V. What about the "missing" 6.5V?


If you have a battery and LED with no current-limiting resistor, then you basically have the same scenario as when you just connect a wire across the two battery leads, but now with only 6.5 V instead of 9 V.

Basically, in this scenario you can no longer neglect the internal resistance of the battery or the forward resistance of the LED (and possibly the resistance of the wire) if you want to determine the actual current. A very large current is produced.


  What happens to the battery, and what to the LED?


Like in the short-circuited battery case, the battery heats up. If the LED is only rated for 20 mA, and you put 100 mA or 1 A through it like you might in this situation, then very quickly the light-emitting diode becomes a smoke-emitting diode (at least momentarily) and then it usually becomes an open circuit.


  What is the actual current in the circuit?


As I mentioned above, you need to consider the internal resistance of the battery and the equivalent resistance of the LED, and maybe the wire resistance. From these you can estimate the actual current. You probably can't calculate it exactly because the numbers depend on temperature, which will be rapidly changing as the circuit heats up and then cools down after the LED pops.