If 258 mL of 2 molar HCl solution is added to 342 mL of 4.3 molar Ba(OH)2 solution, what will be the molarity of BaCl2 in the resulting solution? Answer in units of M.

This is not a very good problem because not all of the Ba(OH)2 is used AND because there is no BaCl2 formed. All of the Ba^2+ and all of the Cl^- is there that you started with; i.e., no actual BaCl2 is formed. So I don't understand what the problem wants since there is no BaCl2 as such.

To find the molarity of BaCl2 in the resulting solution, we need to use the concept of stoichiometry and the balanced chemical equation between HCl and Ba(OH)2.

The balanced chemical equation for the reaction between HCl and Ba(OH)2 is:
2HCl + Ba(OH)2 -> BaCl2 + 2H2O

From the equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 1 mole of BaCl2.

First, let's calculate the moles of HCl and Ba(OH)2 in the given solutions.

Moles of HCl = Volume of HCl solution (in liters) x Molarity of HCl solution
= 0.258 L x 2 M
= 0.516 moles

Moles of Ba(OH)2 = Volume of Ba(OH)2 solution (in liters) x Molarity of Ba(OH)2 solution
= 0.342 L x 4.3 M
= 1.4714 moles

Since the reaction between HCl and Ba(OH)2 is in a 1:2 molar ratio, we need to determine the limiting reactant. The limiting reactant is the one with fewer moles.

In this case, HCl is the limiting reactant because it has fewer moles (0.516 moles) compared to Ba(OH)2 (1.4714 moles).

To find the moles of BaCl2 formed, we can use the stoichiometry of the reaction:

0.516 moles HCl x (1 mole BaCl2 / 2 moles HCl) = 0.258 moles BaCl2

Finally, we can calculate the molarity of BaCl2 in the resulting solution:

Molarity of BaCl2 = Moles of BaCl2 / Volume of resulting solution (in liters)
= 0.258 moles / (0.258 L + 0.342 L)
= 0.451 M

Therefore, the molarity of BaCl2 in the resulting solution is 0.451 M.