A chemist mixes 100ml of H2O at 30 degrees Celsius and 56.0g of CaOin a calorimeter.

Question

The following reaction occurs"
CaO(s) +H2O(l)--->Ca(OH)2(s) ΔH= -16.0 kcal/mol

The heat capacity Ca(OH)2(s) is 19.2 cal/mol*degreeCelsius and the heat of vaporization of water is 540 cal/g. How much liquid water and how much steam would result?

Answer 1

I would do this.
56 g CaO is 1 mole; it will produce 1 mole Ca(OH)2 in the reaction.
The reaction will evolve 16,000 calories for the 1 mole CaO used.
How much will the water use in moving from 30 C to 100 C?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = 100 x 1 x (100-30) = ??approximately 7000 cal.
How much will the Ca(OH)2 absorb?
That will be 1 mole x 19.2 cal/mol x (100-30) = ?? approximately 1400 cal.

Total heat absorbed about 7000+1400 = about 8400 cal. How much heat is left over? 16,000-8400 = about 7600 cal. Where will it go? To vaporize the water to steam. How much will be vaporized.
7600 = mass water vaporized x 540 cal/g and solve for mass water vaporized = about 13 g; therefore, about 13 grams will be steam and 100-13 = ?? will be liquid water at 100 C.