Help with a molarity problem?

If 400. mL of .200 M HCl solution is added to 800.mL of .0400 M Ba(OH)2 solution, the resulting solution will be how much molar in BaCl2?

Answer:
2HCl + Ba(OH)2 - BaCl2 + 2H2O
(0.08moles) (0.04 moles) since ratio is 2 :1
molarity of solution = 0.04 moles/ (0.4L + 0.8L)
= 0.04 / (1.2L)
= 0.03M
First we find the number of moles of each of the reactants
for HCl = 0.4 x 0.2
=0.08 moles
for Ba(OH)2 = 0.8 x 0.04
=0.032 moles.
Now since ratio of Ba(OH)2 to HCl is 1:2, the number of moles of HCl required to neutralize the Barium hydroxide will be = 0.032 x 2
= 0.064 moles.
This means that moles of BaCl2 formed will be
= 0.032 moles.
Now theses many moles of BaCl2 are present in (400 + 800) mL of solution
So the Molarity of BaCl2 in the resulting solution will be:
= 0.032/1.2
=0.02667 M

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