[HelloMotto]

A chemist wants to determine the enthalpy of neutralization for

HCL + NaOH ~> NaCl + H20

There is 61.1 ml and 0.543 mol/L of HCL. There is 42.6 ml of NaOH.
The initial Temperautre is 17.8 celcius and the final is 21.6 celcius. Calculate the enthalpy of neutralization in kJ/mol of HCl. Assume Density for both solution is 1.00g/ml and the specific heat capacity for both is 4.184.

Ok so as my text book said, I totaled the volume of both reactans, and converted that to grams.
103.7g.

And then I used Q=mc(t2-t1) formula to find Q, then Q = -Q solution, which is -1648.74704 J.

I then found the mold of HCL , 0.0331773 mol.

Then finally, I plug thes values into H = -Q/n. So the answer I got was  H = -49.7 kJ.
My textbook dosent have any answers...and its my first time solving this question. Could you guys check if my answer is right?

[Astrokel]

correct, good try!