Calculating Heat Released in Ba(s) + O2(g) → 2BaO(s) Reaction

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The reaction of barium (Ba) with oxygen (O2) releases -1107 kJ for the formation of 2 moles of barium oxide (BaO). For 5.75 g of Ba, which is approximately 0.0412 moles, the heat released is calculated using the reaction's stoichiometry. The correct calculation shows that the heat released is 23.2 kJ, as this value is based on the enthalpy per mole of reaction rather than per mole of Ba. The confusion arises from interpreting the coefficients in the balanced equation, where the enthalpy change applies to the entire reaction rather than just the limiting reactant. Understanding this concept is crucial for accurate thermodynamic calculations.
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The value of H° for the reaction below is -1107 kJ:
2Ba(s) + O2(g) --> 2BaO(s)

How many kJ of heat are released when 5.75 g of Ba(s) reacts completely with oxygen fo form BaO(s)?


i worked out the moles of Ba which was 0.0412 then times by -1107 whcih is 46.4 however the answer is 23.2, i don't understand why becasue the is a 2 to 2 relationship between Ba and BaO? any help would be good




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The specific enthalpy of reaction would be -1107 kJ per 2 moles of Ba consumed (or per mole of oxygen consumed or per 2 moles of BaO produced). This would give the 23.2 kJ of heat released.
 
thanks, but how do u no you have to use the coeffecient of the oxygen, when the Ba is limiting?
 
Value is given not per mole of Ba, but per so called 'mole of reaction'.

That's not always the case.
 
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