how many mL of .135M Sr(OH)2 are needed to neutralize 35ml of .157M HCL

I know the answer came out to 20.4 ml
My only question is how do I convert HCL to Sr(OH)2

3 answers

You've asked two or three questions about how to make the conversions. They're all done with dimensional factors.
Write and balance the equation.
Sr(OH)2 + 2HCl ==> SrCl2 + 2H2O

This equation TELLS you that 1 mol Sr(OH)2 reacts with 2 mols HCl to form 1 mol SrCl2 and 2 mol H2O. So those factors are used to make the conversion. Knowing any one of the four you can calculate all of the others.
mols HCl = M x L = 0.157 x 0.0350.005495
0.005495 mol HCl x (1 mol Sr(OH)2/2 mol HCl) = 0.005495 x 1/2 = 0.0027475

Then M Sr(OH)2 = mols/L
mols = 0.0027475; L = ?; M = 0.135
So L = 0.0027475/0.135 = 0.02035L = 20.35 mL. You are allowed 3 significant figures so I would round this to 20.4 mL.
Hope this helps. Knowing mols HCl allows you to use factors to calculate mols of anything else in the equation.

why is it 2HCL

The equation must be balanced.