[stlboy]

I know molarity essentially is moles of solute/litres of soln

question is taking .417 L of 4 molar HCL adding to .347 L  of 2.6 molar Ba(OH)2

need molarity of BaCL2

I calculated # moles of HCL (.417 * 4 = 1.668)
  and          # moles of Ba(OH)2 (.347 * 2.6 = .9022)

know equation balanced is  Ba(OH)2 + 2HCL ==> BaCL2 = 2H20

then I am stuck, maybe overthinking (but probably underthinking)

I thought i would add moles and divide by total litres   (1.668 + .9022)/(.417 +.347) but that
did not seem right

Any explanations is appreciated,

[Borek]

This is a limiting reagent question.