To calculate the amount of energy released during the reaction of barium hydroxide with nitric acid, we typically need the enthalpy change (\( \Delta H \)) for the reaction. However, this information is not provided in your question.
For the purpose of this calculation, let's assume you have the standard enthalpy change for the reaction (\( \Delta H \)).
Based on the reaction you provided:
\[ \text{Ba(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ba(NO}_3\text{)}_2 + 2\text{H}_2\text{O} \]
The stoichiometry shows that 1 mole of barium hydroxide reacts with 2 moles of nitric acid to produce 1 mole of barium nitrate and 2 moles of water.
If we assume you have a value for the enthalpy change \( \Delta H \) in kJ/mol for the reaction, follow these steps:
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Calculate the number of moles of nitric acid required for the reaction: \[ \text{Moles of HNO}_3 = 2 \times \text{Moles of Ba(OH)}_2 = 2 \times 0.18 = 0.36 \text{ moles} \]
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The total energy released is calculated using the number of moles of barium hydroxide with the enthalpy change: \[ \text{Energy released} = \Delta H \times \text{moles of Ba(OH)}_2 \]
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Substitute in the values:
- If, for example, the reaction has an enthalpy change of \(-57.1\) kJ/mol (this value is hypothetically given; you would need the correct value for your specific reaction): \[ \text{Energy released} = (-57.1 , \text{kJ/mol}) \times 0.18 , \text{mol} = -10.3 , \text{kJ} \]
Conclusion
You will need the correct enthalpy value for the specific reaction to finalize the calculation. Once you have that, simply apply the formula, and you will be able to calculate the total energy released. If you have a specific enthalpy value, please provide it, and I can help you carry out the calculation.
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