If 252mL of 2.5molar HCl solution is added to

Question

343 mL of 2.8 molar Ba(OH)2 solution, what
will be the molarity of BaCl2 in the resulting
solution?

Answer 1

mmols HCl = 252 mL x 2.5M = about 630
mmols Ba(OH)2 = 343 x 2.8 = about 960
but you should confirm all of the calculations Some are estimates.

2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
First, which is the limiting reagent (LR)

Convert mmols HCl to mmols BaCl2. That's 630 x (1 mol BaCl2/2 mol HCl) = about 315 mmols BaCl2.
Convert mmols Ba(OH)2 to mmols BaCl2. That's 960 x (1 mol BaCl2/1 mol Ba(OH)2) = 960 mmols BaCl2.

You see that the two values don't agree (the usual case in LR problems) and one must be wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the LR.
So we have 315 mmols or 0.315 mols BaCl2 formed. The volume of the solution is 252 mL + 343 mL = 595 mL or 0.595 L. Then M = mols/L solution.