Why Divide the Moles of K by 2 in Combustion Calculation?

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SUMMARY

The discussion centers on the calculation of the internal energy of formation for K2O during the combustion of potassium (K) in a calorimeter. The combustion reaction is represented as 2K + 1/2O2 → K2O, indicating that two moles of potassium yield one mole of potassium oxide. Participants clarified that the number of moles of K must be divided by 2 to accurately reflect the stoichiometry of the reaction, ensuring correct calculations of internal energy using the equation DeltaU combustion = -(M/m)(m(H2O)/M(H2O)C(H2O)deltaT + Ccal delta T).

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Samuel1321
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Homework Statement


A sample of K (s) of mass 2.740g undergoes combustion in a constant volume calorimeter at 298.15K . The calorimeter constant is 1849 J/K , and the measured temperature rise in the inner water bath containing 1450 of water is 1.60K
Calculate the internal energy of formation for K2O

Homework Equations


DeltaU combustion = -(M/m)(m(H2O)/M(H2O)C(H2O)deltaT + Ccal delta T)

The Attempt at a Solution



2K+1/2O2 --> K2O

I basically just plugged everything inside but I was off by a factor of 2. When I look at the the answer key, they divided the number of moles of K by 2, I'm not quite sure why do we need to divide the number of moles of K by 2?
 
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How many moles of K2O produced per each mole of K?
 
Borek said:
How many moles of K2O produced per each mole of K?

2, I think I get it now, thanks!
 
Samuel1321 said:
2

Quite the opposite!
 

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