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    How many grams of propane must be burned to supply the heat required to vaporize 0.680 L of water at 298 K?
    C3H8(g) + 5 O2(g)-----> 3 CO2(g) + 4 H2O(l) Change of Heat = -2.22 x 10^3 kJ
    H2O(l) ------> H2O(g) Change of Heat of Vapn. = 44.0 kJ

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    https://brainmass.com/chemistry/energetics-and-thermodynamics/heat-required-vaporize-water-162759

    Solution Preview

    (1) determine the amount of heat required to vaporize 0.680 L of water at 298K. That is obtained from the heat vaporizaton of water given in the problem:
    H2O(l) ------> H2O(g) Change of Heat of Vapn. = 44.0 kJ
    Remember that this is 44.0 kJ ...

    Solution Summary

    It finds the mass of propane that must be burned to supply the heat required to vaporize water at 298 K.

    $2.49

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