Chemistry help ( please 😭😭 )

Hi I’m really struggling with how to answer the first part of this question . Any help would be much appreciated!!!
Equation 22.1 H,SO (aq) + 2NaOH(aq) -› Na,SO (aq) + 2H,0(1)
This is a neutralisation reaction. A student carries out an experiment to determine the enthalpy change AH, and uses this value to deduce the enthalpy change of neutralisation. The student measures out two solutions: 25.0 cm3 of 1.60 moldm-3 H2SO4. 55.0 cm3 of 1.50 moldm-3 NāOH(aq) .The student mixes the two solutions. The temperature increases by 13.0°C.
1) Calculate the enthalpy change of in KJmol-1 ( this is the bit I can’t do )
2) Calculate the enthalpy change of neutralisation in KJmol-1 ( managed to do this but ).
Thank you

Hi! Im not sure if this is right but I think you start by finding the limiting reagent, so if you x the conc by volume for each you get 0.04 moles of sulfuric acid and 0.0825 moles sodium hydroxide. The ratio of moles for the acid to hydroxide is 1:2 so if u do the moles of acid x by 2 you get 0.080 moles of NaOH u need to react with the acid, but because you have 0.0825 moles of NaOH you have more than you need so its in XS and the acid is the limiting reagent. The equation is q=m x c x temp change. The mass would be the total volume, as im assuming the density is 1 g cm3, and mass is density x volume so 25 + 55 =80 cm3 x 1 g cm3 = 80g. So if u subsitute the numbers in you get 80g x 4.81 x 13 = 5000.2 J (which is q). Then enthalpy change (kjmol-1 ) is q divided by 1000 (to get from J to kJ) divided by the limiting moles which was 0.04, so 5000.2/1000/0.04 = -125 kJmol-1 (negative as the temp change was an increase which shows it was exothermic, which is always a negative enthalpy change) . Let me know if the answer was right! 😊