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3] 60.5 ml of HNO3 are required to react with 25.0 ml of a 1.00 m Barium hydroxide solution: HNO3(aq) + Ba(OH)2(aq) --> H2O(s) + Ba(NO3)2(aq) (unbalanced) find the Molarity of the nitric acid solution answer
Accepted Answer
1. Balance the chemical equation:
2 HNO₃(aq) + Ba(OH)₂(aq) → 2 H₂O(l) + Ba(NO₃)₂(aq)
2. Calculate the moles of barium hydroxide:
Moles = Molarity × Volume (in Liters)
Moles of Ba(OH)₂ = 1.00 M × 0.025 L = 0.025 moles
3. Determine the mole ratio:
From the balanced equation, 2 moles of HNO₃ react with 1 mole of Ba(OH)₂.
4. Calculate the moles of nitric acid:
Moles of HNO₃ = 0.025 moles Ba(OH)₂ × (2 moles HNO₃ / 1 mole Ba(OH)₂) = 0.050 moles
5. Calculate the molarity of nitric acid:
Molarity = Moles / Volume (in Liters)
Molarity of HNO₃ = 0.050 moles / 0.0605 L = 0.827 M
2 HNO₃(aq) + Ba(OH)₂(aq) → 2 H₂O(l) + Ba(NO₃)₂(aq)
2. Calculate the moles of barium hydroxide:
Moles = Molarity × Volume (in Liters)
Moles of Ba(OH)₂ = 1.00 M × 0.025 L = 0.025 moles
3. Determine the mole ratio:
From the balanced equation, 2 moles of HNO₃ react with 1 mole of Ba(OH)₂.
4. Calculate the moles of nitric acid:
Moles of HNO₃ = 0.025 moles Ba(OH)₂ × (2 moles HNO₃ / 1 mole Ba(OH)₂) = 0.050 moles
5. Calculate the molarity of nitric acid:
Molarity = Moles / Volume (in Liters)
Molarity of HNO₃ = 0.050 moles / 0.0605 L = 0.827 M
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