Battery for my robot

Hello friends.
I am struggling to get all the parts necessary for the arduino UNO robot I want to make. The robot will have an ultrasonic sensor, an I2C lcd display, few LED's, 2 pushbuttons, a potentiometer and an IR remote. Also it will have the L298N motor driver. The 2 motors are 6-12V and go up to 350 rpm. But I am not sure what battery to get. I think I read somewhere that the battery(battery pack) should have about 12V, but I am not sure. So please, can you tell me what battery should I buy and maybe even send a link to where to buy that battery pack.

Thanks in advance.
Lovro.

The motors will consume almost all of the power, so you need to know how much power they will consume on average, and decide how long you want to run the robot. With that knowledge in hand, then you can shop for batteries.

The L298N is an ancient, inefficient motor driver (it wastes a lot power as heat) and works for only small motors. Post a link to your motors and we can suggest better options.

Hi. The motors I will be using come with the robot chassis I'm going to order. I would make my own chassis, but I want the robot to have tracks. This is the link to the robot chassis and motors:

Also if not L298N, what motor driver do you suggest I use.
Thanks in advance.

There is no useful information on the web page about the motors, except the figure 100 mA, which may be the no load current.

You will have to determine the average current the motoros consume while the tank is driving. Use your multimeter to measure it.

Oh, damn. Ok uh, well thanks for your help. When the damn thing arrives, I'll use the multimeter and I'll know what to do next. But I don't know, L298N looks pretty good to me. I mean, I saw videos of people controling their robots with it and it looked pretty good.

The L298N will probably work with that motor. However, it drops up to 4V internally and wastes the power as heat.

From a 12V power supply, that means the motors get only 8V; about 2/3 of the available power.

Modern motor drivers are far more efficient.