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March 28, 2017

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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

  • Chemistry - ,

    So you get 65.2 kJ heat from 1 mole (74.09 g) Ca(OH)2.
    -65.2 kJ x (273/74.09) = ??

  • Chemistry - ,

    What he said, except you actually multiply by positive 65.2kJ.

  • Chemistry - ,

    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

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