Answer:
Step-by-step explanation:
We'll assume there is an excess of silver nitrate, so that all 12.0 moles of the magnesium (Mg) will react.
The balanced equation tells us we'll obtain 2 moles of Ag for every 1 mole of magnesium, for a molar ratio of 2/1.
Starting with 12.00 moles Mg, we would therefore hope to find twice that, or 24.00 moles of Ag.
To convert to grams, find the molar mass of Ag from the periodic table.
Ag has a molar mass of 107.9 (to 4 sig figs) grams/mole.
(24.00 moles)*(107.9 grams/mole) = 2590 grams (4 sig figs)
Hands off, it's mine.