So you get 65.2 kJ heat from 1 mole (74.09 g) Ca(OH)2.
-65.2 kJ x (273/74.09) = ??
How much heat, in kilojoules, is associated with the production of 273kg of slaked lime, Ca(OH)2?
CaO(s)+H2O(l)-->Ca(OH)2(s)
delta H=-65.2kJ
4 answers
What he said, except you actually multiply by positive 65.2kJ.
In order to solve this, you need to relate the ratio of kJ of heat and the moles of Ca(OH)2 in the reaction, to an unknown kJ of heat and moles of Ca(OH)2 in 273kg. 1 mole of Ca(OH)2 = 74.09g
Set up a proportion:
65.2kJ/74.09g = ??kJ/273000g
Then cross multiply the 65.2 and the 273000, and divide by the 74.09
Set up a proportion:
65.2kJ/74.09g = ??kJ/273000g
Then cross multiply the 65.2 and the 273000, and divide by the 74.09
so what u do is u use pv + nrt to find number of moles then u use q = mc delta t to find the amount of heat. And that should give u the answer. (Don't doubt my knowledge as I am a former member of the 2012 US IChO Team).
1 answer