If a 0.4681 g mg strip reacts with 0.650 m hcl in a 139.3 ml flask at 25oc, what is the minimum volume (ml) of hcl needed to completely react with the magnesium?

The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

    Mg + 2HCl --> H2 + MgCl2

The number of moles of HCl that is needed for the reaction is calculated below.
    n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
    n = 0.0385 mols HCl

From the given concentration, we calculate for the required volume. 
    V = 0.0385 mols HCl/(0.650 mols/L)
     V = 0.05926 L or 59.26 mL

59.26 mL of HCl

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.