How many (mm) of 2.00 M HCL are required to react with 4.55 g Zn?

Answer 1

2HCl + Zn ==> ZnCl₂ + H₂ ... balanced equation

moles Zn present = 4.55 g Zn x 1 mol Zn / 65.4 g = 0.06957 mols Zn

moles HCl needed = 0.06957 mols Zn x 2 mol HCl / mol Zn = 0.1391 mols HCl

The question asks for mm HCl but it probably is meant to ask for ml (millilliters) of HCl:

0.1391 mols HCl x 1 L / 2.00 mols = 0.06957 L

0.06957 L x 1000 mls / L = 69.6 mls