Obviously the style of beer, type of grain (barley, wheat, etc) and depth of roast would make a difference.
But in general what would be the typical ranges be?
So for example:
If you give imperial measurements (lbs Gal) I can covert them no problem.
(Bonus points: And when and why would you deviate from that?)
It will vary, but to give you a benchmark, you can use this Brewer's Friend Calculator to play with the variables of amount of grain, gravity and volume of beer.
For example, if I plug in 100 liters as the volume, 1.050 as the measured gravity (this isn't really important - it just calculates pre-boil efficiency), and choose 25kg of American Pilsner malt, it tells me that this corresponds to 65% efficiency, which is typical for some systems.
So in round figures, 25kg for 100 liters - that's 0.25kg malt per liter.
Just to check if it's in the ballpark, I compare it to my recent brews, I use typically use 8kg for 40 liters of 1.050 SG (ca. 4.3%) beer, which is 0.2kg malt per liter.
This week, I'm going to brew a 14% Imperial Stout, with 25kg of grain for 40l of beer, so that's 0.625kg malt per liter.
The biggest contributors to how much malt you need is the desired starting gravity and brewhouse efficiency. Using different grains don't make a huge amount of difference - those grains with low yields are typically only a small part of the grain bill.
You ask when you'd deviate from this. The amount of grain per volume of beer depends upon many factors:
This glosses over many details, but that's the essence of it - the brewhouse efficiency determines how well you used the grain, so lower efficiency requires more grain, and with a higher starting gravity, you of course need more grain for the given batch size.
I see from your profile you're a developer - I feel you might be trying to be too systematic about this. To know how much grain you need, it's best to follow an existing recipe, one where the mash efficiency is around 70%. That will give both the quantity of grain and the recipe yield.
Your question is like saying "how long is a piece of string?". But here's one way to look at it....on average, at 100 % efficiency, a lb. of base grain in a gal. of water will give you 36 gravity points, or an OG of 1.036. Assuming 75% efficiency, about average for most brewers, you'd get 25 ppg (points/pound/gal.) or an OG of 1.025 for one gallon. So, there is no real standard. You use as much grain as you need to get the OG you want. For instance, assuming 75% efficiency, you'd use 2 lb. of grain per gallon of water for a gallon of 1.050 beer. If you wanted 5 gallons of beer, you'd multiply that by 5 and use 10 lb. of grain and 5 gal. of water. Of course, this doesn't account for grain absorption or wort boil off,
well it depends on the style, so I might make a whole grain ale that is very pale and assuming a good conversion ratio and easy math figure of 1lb per gallon. liter is about 1/4 gallon, so 1/4 pound. round up a little and call that 125g.
of course working with such a small amount you will likely get much worse conversion, and you could probably just double that amount, 250g pale malted barley.
If using Liquid Malt Extract(I use Muntons, there are others available) a good starting point for making a great Home Brew is 1L of malt + 9 L of water = 10 L of wort. This is a GREAT start: it will be a well balanced body. Hops? That's your call. Hops makes your base beautiful.
