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

[Samuel1321]

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?

 

[Borek]

How many moles of K₂O produced per each mole of K?

 

[Samuel1321]

How many moles of K₂O produced per each mole of K?

2, I think I get it now, thanks!

 

[Borek]

2

Quite the opposite!