How do I determine the molarity of Ca(OH)2? Is my answer right

Given equation: Ca(OH)2 (aq) + 2 HCl(aq) --> CaCl2(aq) + 2H2O (l)
25.24 mL of 0.4500 M HCl was required to neutralize 30.00mL of Ca(OH)2 soln. Determine the molarity of Ca(OH)2.

Here is what I did: (converted mL to L 1st)
0.02525L * 0.4500mol HCl/1 L HCL * 1 mole Ca(OH)2 / 2 mole HCl = 0.005681 moles Ca(OH)2
Molarity = moles/L, so,
0.005681 moles Ca(OH)2/ 0.03000L Ca(OH)2 = 0.1894 mol/L
Is this correct?

is it 25.24 or 25.25 mL?
and you should carry one extra sig fig during intermediate steps. Doesn't matter in this case but it's a good habit to get into.

first conversion
25.25mL x (1L/1000mL) x (0.4500mol HCl / L) x (1mol Ca(OH)2 / 2mol HCl) = 0.0056812mol Ca(OH)2

then.
(0.0056812mol Ca(OH)2 / 30.00mL) x (1000mL / 1L) = 0.1894M