What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?

The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl > CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol  in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol  / 0.250 mol/L = 40.0 mL 
40.0 mL of 0.250 M HCl is required

  super sorry but i dont know lo

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