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
