Spotting the booster rocket of the Chinese space station? SpaceWeather.com's newsletter states

Sky watchers are reporting two new objects in the night sky: China's new space station and an out-of-control booster rocket that helped launch it. The tumbling booster is flashing brighter than some of the brightest stars in the sky.

My usual star-gazing app does not have the rocket booster in its database (yet), and I did not manage to spot it yet (from cloudy central Europe). Is there an official name for the object? Could somebody help me how to find it in the remaining 2 days while it is still in orbit around Earth? Or maybe somebody could even share a picture or two?
Bonus question: Maybe somebody could also enlighten me in how far any single-use booster-rocket could be in-control after it ejected its payload? Wasn't that the default for many years? Also: Now that we can observe the rocket we know the trajectory and could also forecast when it will crash and where, aren't we?
References

U.S. Space Command tracks Chinese rocket for uncontrolled re-entry from orbit
The name of the module is Tianhe (天和), and the name of the rocket is Long March 5B, but it got a special designation of 2021-035B. It is not really official, but I think it is good enough to get some data. I found this website for current position (live on globe). Your userpage says you live in Europe, so the rocket will fly over Greece at 00:30 CEST (22:30 UTC) tomorrow (in 1.5 hour if you have time to be awake for that long). I also calculated the maximum angular speed (on May 6) of around 1.85 °/s (on May 7, it is 2.7 °/s).
There are already some images; I found only that one. The booster's visual magnitude is around –1, so pretty bright to spot.
So, if you have time, try to look to the South and spot the booster at 22:30 UTC today. Maybe you are going to have luck tonight : )

There are some equations I derived with basic trigonometry about the rocket's visibility:
Distance $D$ at which the rocket can be seen right on horizon:
$$D = R \ \arccos\left(\frac{R}{R+h}\right)$$
Where the calculator calculates radians, $R$ is Earth's radius and $h$ is the altitude of the booster (you can get that on previously given website; at the moment it is 170 km).
The angular height of the booster which is $D$ kilometers away from us (on Earth) is given by:
$$\alpha = \frac{\pi}{2}-\arcsin \, \left(\frac{(R+h)\sin{\frac{D}{R}}}{\sqrt{2R^2+2Rh+h^2-2R(R+h)\cos{\frac{D}{R}}}}\right)$$
Again, the calculator must be set in radians.

Update: The rocket fell on May 9 at 02:14 UTC into the Indian Ocean.