Question

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

Answer 1

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