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91.2 mL of 0.120 M HCl is mixed with 50.0 mL of 0.101 M Ba(OH)2 solution....

91.2 mL of 0.120 M HCl is mixed with 50.0 mL of 0.101 M Ba(OH)2 solution.

2 HCl(aq) + 1 Ba(OH)2(aq) →   1 BaCl2(aq) + 2 H2O(l)

What amount of hydronium ion (in mole) would be present at the end of the reaction?

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