If 20.0g of magnesium react with excess hydrochloric acid how many grams of magnesium chloride are produced?

Well, to start, the equation of the reaction is Mg + 2HCl = MgCl2 + 2(H+)
so the molar ratio of Mg: MgCl2 is 1:1

so we multiply the grams of magnesium by the inverse molar mass to get moles of magnesium:

20.0 g Mg * (1 mol / 24.31 g) = 0.8227 mol Mg

then since the molar ratio is 1:1 that translates directly into moles of MgCl2:

0.8227 mol Mg = 0.8227 mol MgCl2

and finally multiply by the molar mass of magnesium chloride to get the number of grams:

0.8227 mol MgCl2 * (95.21 g / 1 mol) = 78.3 g MgCl2 (only 3 significant figures because the initial 20.0 g only has 3)