How much water for each kg of grains is typically used in mash? (I am referring to mash only, not to sparging). Is that amount the amount of the final wort I will be boiling?
There's a very good explanation of the whole detailed calculations here
...but the sum of that is basically this:
So that would be about 2.6 L per 1 kg (thanks @Paolo for the correction)
You will then sparge with enough water to get it to your desired volume, plus another litre or so to account for evaporation (it's easiest to just know what line on your kettle to fill it to).
Hope that helps :)
There are three factors to consider:
A -amount remaining in kettle after boil B -amount lost to boil C -amount absorbed by grains
For a 90 minute boil "B" will generally be 1.5 gallons.
For "C" I calculate .06 gallons per pound of grains.
"A" is dependent on your batch size.
There are a lot of factors that affect the amount of water used for mashing: batch size, evaporation rate, equipment loss, grain absorption, etc. You're best to start with the amount of beer you want to make and work backwards from there. There are plenty of online calculators that will do the math for you.
If you don't know the values for some of the variables (like "trub loss", which is wort left behind in the boil kettle to avoid adding the trub to the fermenter), just use the default values in the calculator. On your brew-day measure everything and enter those values into the calculator next time. You're probably going to miss your target volume the first few brews, but if you'll eventually dial it in.
For batch size, you might want to enter a larger volume than what you plan to bottle or keg. Most brewers loose close to half a gallon of beer due to yeast and trub sediment in the fermenter.
I'm not sure from the wording of your question if you're asking about "no sparge" mashing. If you are, you can still use the linked calculator. Just keep increasing the "mash thickness" parameter until the "sparge water" result is very close to 0.