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

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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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