Chemistry Heat Capacity Question

The standard enthalpy of evaporation of CF2Cl2 is 17.4 kJ/mol, compared with deltaH of vap= 41 kJ/mol for liquid water. How many grams of liquid CF2Cl2 are needed to cool 128.4 g of water from 40.9 to 26.9 degrees C? The specific heat of water is 4.184 J/(g * degrees C).

Hints:
1. CF2Cl2 is evaporated to cool 128.4 g of water 14.0 degrees C

2. The heat lost by water is given by:
q(water)= m(cs)deltaT
where m=mass, cs = specific heat of water, and deltaT= the difference in temperature

3. The heat gained by the CF2Cl2 to evaporate is:
q(vap.)=n(deltaH of vap.)
where n=moles

4. We do not need to use the deltaH of vap. of water to solve this problem, only the deltaH of vap. of CF2Cl2

5. Incorrect answers: -523 g and -5.23x10^7 g

hi...can you please tell me why my answer was wrong for this question ...
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are you sure that the question what you have typed is correct ?
please i really need to know what is wrong in that ...

now for this question...

first calculate heat lost by water

q = 128.4 X 4.184 X (26.9-40.9) = 128.4 X 4.184 X -14 = -7521.158 j (-ve sign because heat is released

now this much heat will be used by CF2Cl2 to get converted into vapour form

1 mole of CF2Cl2 = 120 g

when there is 17.4 kj or 17400 j of heat than amount of CF2Cl2 evaporated = 1 mole or 120 g

so when there is 1 j of heat than amount of CF2Cl2 evaporated = 120/17400 g

so when there is 7521.158 j than amount of CF2Cl2 evaporated = (120/17400) X 7521.158 = 51.87 g

hope i am correct this time ....also please tell me about that question