Running one of the motors at 24V, unloaded, will draw 4.8A. This will have to continue for 2h30m (2.5h). The required battery capacity for one motor will be:
$$ Q = 4.8A \times 2.5h = 12Ah $$
Note: The units are consistent, \$A \times h = Ah\$. Voltage is not considered here, since then \$V\$ would also have to show up in the answer's units somewhere.
The power rating of the motor is 150W at full load. This will give you some idea of the current that the motor will draw under maximum mechanical load:
$$ I_{MAX} = \frac{P}{V} = \frac{150W}{24V} = 6.3A $$
I have assumed that this power rating is electrical input power, not mechanical output power. If this power is mechanical output power, then you must account for motor efficiency to find the corresponding electrical input power. Refer to the motor's documentation.
At full load, then, the required battery capacity will be:
$$ Q_{MAX} = 6.3A \times 2.5h = 16Ah $$
I have assumed that the batteries are 24V. If each battery pack is only 12V, you must connect two such batteries in series, for 24V. Doing so will not change the capacity (in Ah), since being in series they both pass the same current. For instance, with two 6Ah, 12V batteries, they can still pass 6A, and each can still maintain this condition for 1h.
However, at 24V, each Ampere of current delivers 24W of power, in contrast to the 12W per Ampere available from a 12V source. Therefore, even though it seems counter-intuitive that combining batteries in series doesn't change overall charge capacity, it does increase the amount of energy available per unit of charge.
To obtain 16Ah of capacity, from batteries of only 6Ah capacity, they must be connected in parallel, such that the total 6.3A is shared between them. In the next diagram, I am treating each pair of 12V batteries as a single 24V battery. Therefore, there are \$N=3\$ 24V batteries here:
simulate this circuit – Schematic created using CircuitLab
With \$N\$ individual batteries of \$Q_{IND}[Ah]\$ capacity, connected in parallel, you have total capacity:
$$ Q_{TOT} = N \times Q_{IND} $$
or
$$ N = \frac{Q_{TOT}}{Q_{IND}} = \frac{16Ah}{6Ah} = 2.7 $$
You can't have a fractional number of batteries, so round that up to \$N=3\$.
With the motor running under no load, which we calculated above would need 12Ah of battery capacity to keep going for 2.5h, you have:
$$ N = \frac{12Ah}{6Ah} = 2.0 $$
You should still probably use \$N=3\$ batteries, since their capacity will reduce with age and discharge/recharge cycles.
As a sanity check, we can find the longevity of this setup in terms of energy. Start by calculating the total energy contained in an individual fully charged battery. You'll notice there we do involve voltage now. While it wasn't strictly necessary to consider voltage before (it was sufficient to deal with charge, currents and time), if we wish to make our calculations in terms of energy and power, then voltage must enter the equations.
If you are using six 12V 6Ah batteries, each fully charged battery stores:
$$ E_{IND}[Wh] = Q_{IND} \times V = 6Ah \times 12V = 72Wh $$
The total amount of energy available from all six batteries is:
$$ E_{TOT} = E_{IND} \times N = 72Wh \times 6 = 430Wh $$
If you are using three 24V 6Ah batteies, you'll find the energy stored is the same:
$$ E_{IND}[Wh] = Q_{IND} \times V = 6Ah \times 24V = 144Wh $$
The total amount of energy available from all three batteries is:
$$ E_{TOT} = E_{IND} \times N = 144Wh \times 3 = 430Wh $$
Convert those 430Wh to Joules, to be compatible with our power calculations below (because the units of power are joules per second \$[W] = \left[\frac{J}{s}\right]\$:
$$ E_{TOT[J]} = E_{TOT[Wh]} \times 3600 = 1.5MJ $$
Drawing 150W (that's 150 joules per second) from this energy source, we can expect the source to be depleted after time \$T\$:
$$ T = \frac{1.5MJ}{150W} = 10000s \approx 160min \approx 2.8h $$
All that is for one motor. To obtain the same run-time for two motors, you need twice as much energy stored, which means twice as many batteries:
simulate this circuit
or:
simulate this circuit
In terms of charge capacity, this works out as follows: each 24V battery has 6Ah capacity. Connecting six such batteries in parallel increases total capacity:
$$ Q_{TOT} = N \times Q_{IND} = 6 \times 6Ah = 36Ah $$
At a load of \$I=2\times 6.3A = 12.6A\$, this compound battery will last:
$$ T = \frac{Q_{TOT}}{I} = \frac{36Ah}{12.6A} \approx 2.8h $$
Alternatively use two 12V car batteries in series, each with a capacity of over 50Ah. Even a pair of small 50Ah batteries gets you a run time of:
$$ T = \frac{50Ah}{2 \times 6.3A} \approx 4h $$