What volume of 0.109 M nitric acid is required to neutralise 2.50g barium hydroxide?

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asked by Ross
  1. Since the reaction is

    2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2H2O

    It takes 2 moles of HNO3 to neutralize 1 mole of Ba(OH2)

    2.5g Ba(OH)2 is 0.0146 moles
    So, you need 0.0292 moles of HNO3

    0.0292M / 0.109M/L = 0.268L of acid.

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    posted by Steve

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