The correct answer is $\boxed{\text{C}}$.
The rimpull is the maximum tractive force that can be exerted by the drive wheels of a tractor without slipping. It is calculated by multiplying the coefficient of traction by the weight on the drive wheels. In this case, the coefficient of traction is 0.5 and the weight on the drive wheels is 25,000 kg, so the rimpull is 25,000 kg $\times$ 0.5 = 12,500 kg.
The other options are incorrect because they do not take into account the coefficient of traction. Option A is the total weight of the tractor, but this is not the weight on the drive wheels. Option B is the total loaded weight of the tractor, but this includes the weight of the scraper wheels, which do not contribute to the rimpull. Option D is the weight on the scraper wheels, but this is not the weight on the drive wheels.