Stoichiometry help?? Converting volume to molarity.

315 mL of strontium hydroxide solution is completely neutralized with 525 mL of 0.80 M HCL acid solution according to the balanced equation shown below. Calculate the molarity of the Sr(OH)2 solution.
Sr(OH)2+2HCL=SrCl2+2H2O

Determine the maximum mass in grams of barium sulfate that can be formed by the reaction of 245.0 mL of 0.125 M barium chloride and 185.0 mL of 0.240 M sodium sulfate.

You don't have to do both
Thanks in advance!

For the first problem, (0.525 x 0.8) mole HCl will require 0.21 mole Sr(OH)2 assuming complete neutralization.

To get the molarity of Sr(OH)2, divide its number of moles by the volume of its solution in liters.

For the second problem, you have:

(0.245 x 0.125) mole = 0.0306 mole barium chloride

(0.185.0 mL x 0.240) mole = 0.0444 mole sodium sulfate

Assuming that all of the available reactants will react,

0.0306 mol BaCl2 will require 0.0306 mol Na2SO4

0.0444 mol Na2SO4 will require 0.0444 mol BaCl2

From the calculated values above, we can see that we cannot use all of the available Na2SO4 since the amount of available BaCl2 is insufficient.

BaCl2 is the limiting reactant in this problem.

0.0306 mole (6.377 g) BaCl2 will require 0.0306 mole (4.350 g) Na2SO4 to produce 0.0306 mole (7.147 g) BaSO4 and 0.0612 mole (3.580 g) NaCl.