Edit
You provided a link to some technical data, which are:
Weight: 100 gm(Approx.)
RPM: 100
Operating Voltage: 12V DC
Gearbox: Attached Plastic (spur)Gearbox
Shaft diameter: 6mm
Shaft Length: 21 mm
Torque: 1.2 Kg-cm.
No-load current = 60 mA(Max)
Load current = 300 mA(Max).
Dimensions: Body Diameter: 38mm; Motor Length with shaft: 77 mm;
Same size motor available in various RPMs.
As we can see, the motor already has some gearbox attached to it, which reduces the maximum rotational to 100 rpm (=rotations per minute) and increases the available torque at the same time. The maximum torque is given as 1.2 kg*cm (which includes the gearbox-transition), which means having wheels with an radius of 1 cm (or 2cm in diameter) will give you a force that is equivalent to 1.2 kg in weight at nominal voltage, as long as your electronics can provide a current of 300 mA (load current).
However, 1.2 kg "weight" force doesn't mean, that you are limited to a weight of 1.2 kg for e.g. a robot, as 1.2 kg do not mean the weight of the "thing" which is moved but an equivalent force. It means a force that is equivalent to the force that gravity causes at 1.2 liters of water (in container which we asume to have no weight) to be pulled downwards. If you wanted to pull 1.2 kg in upwards direction, you would need a bit more than this force...
If you just need to drive a robot on a 100% horizontal surface you need to overcome Stiction (see formulae above), which for a robot with a weight of 1.2kg on wheels is much less than a force of 1.2kg "weight" (=1.2kg * 9.81m/s² = 11.8N). Anyhow, your robot will also need to climb slightly upwards.
Lets assume 20% inclination, which means you need a force equivalent to sin(20°) times the weight of the robot. And let us assume this force needed is much more than Stiction. In that case you can pull 1/sin(20°)=2.92 times 1.2 kg = ~3.5 kg upwards on a surface with an inclination below 20° with this motor using wheels with an radius of r=1cm. If your wheels are bigger, the maximum robot weight will be less (e.g. r=2cm ==> 3.5kg/2 = 1.75kg).
However, your speed will get near to zero if you inclination is near to 20%. So every weight that is less than this limit ensures you some reserve for accelerating your robot, even at 20% inclination. If our maximum robot weight is 3.5 kg according to the above calculation and the real robot has a weight of 2kg, then your acceleration force will be like 1.5kg "weight" pulling it forward on a horizontal surface. As stated above, 1.5 kg "weight" means in fact a force of 1.5 kg * 9.81 m/s² = 14.715 kg*m/s². Our example robot having a weight m=2kg will be accelerated by a force of F=14.716 kgm/s². The formula that relates mass, force and acceleration is: F = m*a, where F is the force, m is the mass and a is the acceleration. So a = F/m would be 14.715 kg*m/s² divided by 2 kg and therefore: ~7.358 m/s². This means within 1 second your robot would reach a speed of 7.36 m/s (or 26.5 km/h) if there was no friction. Also the maximum speed is 100 rpm; so having the wheel radius of r=1cm the robot will move forward U=2r*3.14=6.28 cm. Having 100 revolutions per minute gives us 628 cm = 6.28 m per minute or 0.1046 m/s as our top speed, if there is just the friction of the motor itself within its gearbox and so on... so it moves 10,46 cm forward per second. If you need more speed you need bigger wheels or some gears... but as said, your maximum weight will drop then.
IMPORTANT: Calculation of all the friction correct is even more complicated, so dont expect the full 10,46 cm per second! This is just an upper limit that is known for sure...
