Asked by Ross
What volume of 0.109 M nitric acid is required to neutralise 2.50g barium hydroxide?
Answers
Answered by
Steve
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.
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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