Calculate the volume of hydrogen (t = 0 °C, p = 101,3 kPa) released from 16,35 g of zinc in excess of HCl

My answer is wrong and i'm not sure why it is, so if you spot my mistake please tell me.
2Zn + 2HCl -----> 2ZnCl + H2

Mr of Zn = 65.38
Number of moles of Zn = 16.35 / 65.38 = 0.2500076


Stoichiometric ratios of the reaction:
2mol Zn ------> 1 mol H2
So number of mols of H2 = 0.250076/ 2 = 0.125038238

volume of H2 = number of mols of H2 * 22.4
= 0.1250... * 22.4 = 2.80dm^3

The answer booklet says the answer is 5.60dm^3. Can you please explain how this answer is derived

the equation for the reaction is
Zn + 2HCl -----> Zn(II)Cl2 + H2 !!

13.35 g Zn = 0.25 moles ---> give 0.25 moles H2

1 mole H2 = 22.4 L
0.25 moles = 22.4/4 = 5.6 L


OG