What is the molarity of chloride ion present in the solution that is formed when 238.50mL of 1.35M barium hydroxide reacts with 2146.5 mL of 0.3 M...

the amount of moles =
volume(ml)*molarity*10⁻³

the amount of HCl
moles = 2146.5*0.3*10⁻³ ~ 0.64mol

the amount of BaCl₂moles
= 238.5*1.35*10⁻³
~ 0.32mol

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Ba(OH)₂+
2HCl → BaCl₂ + 2H₂O

0.32mol  +  0.64 mol →
0.32mol + 0.64
mol

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BaCl₂→
0.32mol 

BaCl₂→ Ba²⁺ +
2Cl

0.32mol  → 0.32mol +
0.64mol

-----------------------------------------------

Cl
→ 0.64mol

2146.5ml + 238.5ml →
0.64mol

2385ml → 0.64mol

∴
1000ml → 0.64*1000/2385 mol ~ 0.268mol

Hence
Molarity = 0.268M