Help with Heat of Neutralization Homework

[sp3sp2sp]

Homework Statement

Im confused about this and need help
We did heat of nuetralization lab where we reacted 25mL of HCL and 25mL of NaOH using calorimeter.
The temp difference was an increase of exactly 7degC
One question asks me to show q_water (J) and to assume 50g water for the calculation
so i plugged in 50g * 4.186J/gdegC * 7degC = 1465.1J

Next it asks me what delta H_nuetralization (J) is and I wrote:
delta H_neut = q_rxn = -q_water = -1465.1J

Next it asks me for net ionic. I wrote
H^(aq) + + OH^-(aq) --> H2O(l)

Then it asks me for delta H_neut (kJ/mol), per mole of water formed and this is where I am confused, because internet search shows value should be about -58kJ/mol

50g water = 2.78mol
so if 50g water produces -1465.1J, then 1 mol water would produce 1465.1J / 2.775mol =527.0J
= 0.527kJmol - this is way off so I must be doing something wrong somewhere.
thanks for any help

Homework Equations

The Attempt at a Solution

 

[Borek]

You didn't produce 50 g of water.

Were you given concentrations of NaOH and HCl?

 

[sp3sp2sp]

Yes the whole experiement was nuetralization reaction. I had 25.0mL of NaOH and 25.0mL of HCl. I understand that the acid and base nuetralize each other and this is exothermic so heat given off but I am still confused about the whole concept for the enthalpy. thanks for any more help

 

[mjc123]

ΔH = -58 kJ/mol means that 58 kJ of heat are released when 1 mole of H⁺ reacts with 1 mole of OH^- to give 1 mole of water. This is what the question means when it asks for ΔH per mole of water formed - not the water initially present in the solutions. You need to work out the number of moles of water formed in the neutralization reaction, and divide 1465.1 J by that number. For that, you need the concentrations of the solutions, which Borek asked you for but you didn't give.