If HCl+NaOh--> NaCl+ H20, How many moles of NaOH will you need to complete the neutralize the HCl

That all depends on the number of moles of HCl. A look at the balanced equation shows that the mole ratio for HCl and NaOH is 1:1, so whatever the moles of HCl is, that will be the moles of NaOH.

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Based on the 1:1 mole ratio, if 0.0105 moles of HCl (which we calculated for you earlier) react, then 0.0105 mols of NaOH will also react.

Use the unit factor method again...
0.0105 mol NaOH x (1 L / 0.0400 mol NaOH) x (1000 mL / 1 L) = 262.5 mL NaOH solution.

Are you sure you mean the concentration to be 0.0400M, and not 0.400M? 0.400M would be more in keeping with your earlier question.
0.0105 mol NaOH x (1 L / 0.400 mol NaOH) x (1000 mL / 1 L) = 26.25 mL NaOH solution.

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Check the strength of the acid and base. NaOH is a strong base and HCl is a strong acid, so they will neutralize in a ratio of 1:1.

That being said, you only need 1 mole of NaOH to neutralize 1 mole of HCl.

Edit: Don't listen to Melanie. If there were a weak acid or a weak base instead, you'd need to do an equilibrium calculation. It is much more important to look at the acid/base strength.

Edit2: Do you want work?

0.0105 mol H(Cl) * (1 mol OH / 1 mol H) = 0.0105 mol (Na)OH

Well , the equation is balanced, so if you have 1 mol of NaOH, it will neutralize 1 mol of HCL, this is a trivial stoichiometry question