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Thermometric Titration Help!!!

I'm in need of some help with calculations for a thermometric titration, I'm going to list the equations I know and the values I have, but I don't have a clue how to calculate the enthalpy change per moles of acid for each neutralisation.
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Specific heat capacity of water = 4.2 J deg-1 g-1

Specific heat capacity of Solutions = 4.2 (same as water)

n=CxV/1000

Q=mc(Delta)T

[Naoh] = 0.994 M
[HCL] = 0.994 M
[HNO3] = 0.990 M
[H2SO4] = 0.50 M
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Naoh + HCL ---> NaCl + H20
Ratio: 1 : 1 1 : 1

HCL
Temperature rise at neutalisation = 6.625 C
Volume of NaOH at neutalisation = 23.00 cm3

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Naoh + HNO3 ---> NaNO3 + H20
Ratio: 1 : 1 1 : 1

HNO3
Temperature rise at neutalisation = 7.125 C
Volume of NaOH at neutalisation = 23.50 cm3

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2Naoh + H2SO4 ---> Na2SO4 + 2H20
Ratio: 2 : 1 1 : 2

H2SO4
Temperature rise at neutalisation = 7.875
Volume of NaOH at neutalisation = 21.50 cm3

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I know its alot of work, but I really don't want to fail Chemistry =( I need to know:

Enthalpy change of formation per mole pf H20 in KJ mol -1 = ?
Mass of Solutions = ?

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Thank you very much in advance, I appreciate any help I can get. =)

-Matt

=( I still no reply

just calculate energy from E=mcdeltaT

and then divide by the number of moles of water formed in the reaction.

example:
(you haven't stated the volume of acid used so I will assume that it is 25cm3)
NaOH + HNO3 ---> NaNO3 + H20
Ratio: 1 : 1 1 : 1

HNO3
Temperature rise at neutalisation = 7.125 C
Volume of NaOH at neutalisation = 23.50 cm3

m=(25+23.5)/1000 kg
c=4.2
deltaT = 7.125

Therefore energy = 1.451kJ

moles of water formed = moles of NaOH reacting = 0.0235 x 0.994 =0.0234

therefore energy per mole = 62.1 kJ mol-1