How much 0.100 m hcl is required to completely neutralize 20.0 ml of 0.250 m naoh?

The balanced equation for the reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of NaOH moles reacted - 0.250 mol/L x 0.0200 L = 0.005 mol 
therefore number of HCl moles reacted - 0.005 mol 
molarity of given HCl solution - 0.100 M
number of HCl moles in 1 L - 0.100 mol 
then 0.005 mol of HCl in - 0.005 mol / 0.100 mol/L = 50.0 mL
volume of HCl required is 50.0 mL

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