In the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCl in an excess of Mg

Since the Mg is in excess, therefore HCl will be fully consumed in the reaction.

The first step is to find the amount of HCl in mol

Let N (HCl) = amount of HCl in mol

N (HCl) = (6 mol HCL/L solution) * (125 mL) * (1 L/1000 mL) = 0.75 mol of HCl

Through stoichiometry

N (H2) = 0.75 mol HCl * (1 mol H2 / 2 mol of HCl)

N (H2) = 0.375 mol H2

Since we are asked for the number of grams of H2 (mass), we multiply this with the molar mass of hydrogen

M (H2) = 0.375 mol H2 (2 g H2 / 1 mol H2)

M (H2) = 0.75 g H2