First calculate the initial moles of HCl.
HCl moles i = (1.45 moles / L) * 0.5 L = 0.725 moles HCl
Then calculate how much HCl is consumed by stoichiometry:
HCl moles consumed = (12.7 g Zn / 65.38 g / mol) * (2 moles HCl / 1 mole Zn) = 0.388 moles
So HCl moles left is:
HCl moles f = 0.725 – 0.388 = 0.3365 moles
For every 1 mole HCl there is 1 mole H, therefore:
H moles final = 0.3365 moles
So final concentration of H ions is:
Concentration H = 0.3365 moles / 0.5 L = 0.673 M