500 mL of .45 M HCL is reacted with 250 mL of .90 M NaOH in a constant pressure calorimeter. the initial temperature of the solution is 21.75°C. The enthalpy of neutralization is -56.2 kJ/mol and the specific heat of water is 4.184 J/g*°C. What is the final temperature of the solution.
So far I have:
HCL+NaOH--H2O+NaCl
-.500L HCL*(.45molHCL/1L HCL)=.225mol HCL
-.250L NaOH*(.90mol/1L)= .225 mol NaOH
-Both are the same with no excess
-(.225)(-56.2)= -12.645kJ
-(-12645J)+4.18(250mL+500mL)+4.184ΔT=0
-4.18(750mL)+4.184ΔT=12645J

Um, is this right so far??? Can anyone help, please?

Question

500 mL of .45 M HCL is reacted with 250 mL of .90 M NaOH in a constant pressure calorimeter. the initial temperature of the solution is 21.75°C. The enthalpy of neutralization is -56.2 kJ/mol and the specific heat of water is 4.184 J/g*°C. What is the final temperature of the solution.
So far I have:
HCL+NaOH-->H2O+NaCl
-.500L HCL*(.45molHCL/1L HCL)=.225mol HCL
-.250L NaOH*(.90mol/1L)= .225 mol NaOH
-Both are the same with no excess
-(.225)(-56.2)= -12.645kJ
-(-12645J)+4.18(250mL+500mL)+4.184ΔT=0
-4.18(750mL)+4.184ΔT=12645J

Um, is this right so far??? Can anyone help, please?