Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 6.00 M HCl(aq) are required to react with 7.85 g Zn(s)?

Question

My answer was 15.3, Im not sure what I did wrong.

Answer 1

7.85 grams zn * 1 mol/65.38 grams * 2 mol HCl/ 1 mol zn = 0.24 mole HCL

Molarity = Mole/Volume

6 = 0.24/Volume

Volume = 0.04 Liters = 40 Milliliters

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Answer 2

Zn(s) + 2HCl(aq) ⟶ ZnCl2(aq) + H2(g)

moles Zn present = 7.85 g Zn x 1 mol Zn / 65.4 g = 0.120 mols Zn

moles HCl needed = 0.120 mols Zn x 2 mols HCl / mol Zn = 0.240 mols HCl

volume of 6.00 M HCl needed = 0.240 mols HCl x 1 L / 6.00 mols = 0.4000 L = 40.0 mls