How many ml of 1.50 m hcl are needed to titrate 25.0 ml of 1.20 m ca(oh)2?

The balanced chemical equation for the above reaction is as follows;
Ca(OH)₂ + 2HCl --> CaCl₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HCl is 1:2
The number of Ca(OH)₂ moles required - 1.2 M /1000 mL/L x 25.0 mL = 0.03 mol
number of HCl moles required for complete neutralisation - 0.03 x 2 = 0.06 mol
the molarity of HCl solution - 1.50 M
There are 1.50 mol in 1 L solution 
Therefore 0.06 mol of HCl in - 1/1.50 x 0.06 = 0.04 L
therefore volume of HCl required is 40 mL 

It is 4.92 e-4 sorry if i am wrong

It shows that the experiment is designed properly.