12v battery banks and amp hours

The setup in your answer is a bad one.

First you parallel two batteries, which you shouldn't, because their voltages are never exactly the same. Their low internal resistance will cause a current from one battery to the other. So feed three LEDs from 1 battery, and the other three from the other.

Then you place all LEDs parallel, which means that you lose (12 V - 3.2 V) x 20 mA = 176 mW per LED in its series resistor, while the LED itself uses only 64 mW. Total power loss is more than 1 W. That's because the large voltage difference between battery and LED. The best way to get a longer endurance is to keep losses in the series resistors as low as possible. So better place two times three LEDs in series, so that their total current is 40 mA instead of 120 mA. The power loss in the series resistors is then (12 V - 3 x 3.2 V) x 20 mA = 45 mW per 3 LEDs, or 96 mW in total. That's less than 10 % of the power loss for all LEDs in parallel.

Then your batteries will last 100 mAh / 40 mA = 2.5 hours or 150 minutes. This is pretty optimal. The batteries' capacity is 1200 mWh, and the LEDs consume 384 mW, so with an ideal conversion you can get a little over 3 hours out of them. But the most efficient conversion using a switching current regulator will get you maybe 85 % efficiency, and then you only gain 9 extra minutes.

edit re comments
An alkaline battery's voltage quickly drops by 10-15 %, and then remains more constant for a great part of the discharge cycle. So either you calculate the resistors for a larger current at the start, and 20 mA for the rest, or for 20 mA at the start, and a lower current later on. The latter solution will give you a longer battery life, but a bit less brightness.

jippie suggests to use a switcher anyway to get more out of the batteries, and it's a thought. You'll have to place the batteries in series to get 24 V to allow a voltage drop as high as possible. The larger Vin/Vout ratio of the switcher will make it less efficient, but overall you should get some extra time from the batteries.

Let's clear up some terminology first. An Amp is not a discrete quantity of energy. It is actually a quantity of electrons per time. Therefore, you don't say "Amps per hour". I like to use the analogy that Amps of current is like speed in your car. If you drove your car for an hour at 60mph (pardon my English unit assumption) and someone asked you about your drive, you wouldn't say "I drove 60mph per hour". That would be silly. Instead you would say "I drove 60mph for an hour."

It's the same with current. A battery doesn't supply 5 Amps per hour, it provides 5 Amps for an hour. Furthermore, a battery that has supplied 5 Amps for an hour has provided 5 Amp-hours (5Ah) of charge. An Amp-hour is an actual discrete quantity. It represents the capacity of the battery.

So you need a battery that can provide 5 Amps for 2 hours at 12V. That's 10 Amp-hours total. More specifically, since it's 12V: $$5A*2hours*12V = 120Wh$$
Of course, you would never want to buy a battery that can only provide exactly the capacity you need. You want some capacity in reserve. So let's double it and say you probably want a 240Wh battery. That way, you'll only use 50% of the capacity in one session.

The Li-ion battery you mentioned in your question has a 9800mAh capacity at 12V. That works out to 117.6Wh. Almost enough to last for two hours with a 5A current draw. As long as you understand what you're doing with a Li-ion battery, that should work, but I would recommend putting two in parallel to double the capacity. When you put two batteries in parallel, their voltage stays the same (12V) but the total capacity doubles (9800mAh * 2 = 19.6Ah).

EDIT: Looking at the website for that Li-ion battery, it is not clear what its continuous current capability is. It may not be able to supply 5A of current continuously without overheating. Without a datasheet or more technical specs, it's probably not safe to use.

One other note, batteries are not perfect energy sources. As you discharge them, their voltage lowers, their internal resistance increases, and their ability to supply current decreases. The more current you draw at a time, the more these negative effects occur. What all that means is a 10Ah battery may not actually supply a full 10Ah before the battery is flat. This is another reason to get a battery that has a larger capacity than the math dictates.