[slu1986]

When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J g−1 °C−1 and that the density of the solution is 1.00 g mL−1 Calculate the enthalpy change, ΔH, for the reaction:

 

                          HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

 The answer is -54.3 kJ, but I have no idea how to set up the equation and get that answer.  If someone could please help guide me in the right direction, I would appreciate it! Thanks 

[Dan]

I'm not suprised you're having trouble, the question is criminally misleading in my opinion.

If I rewrite it (changes are in red) can you solve it now?

When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J g⁻¹ °C⁻¹ mol^-1 and that the density of the solution is 1.00 g mL⁻¹ Calculate the molar enthalpy change, ΔH, for the reaction:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

 The answer is -54.3 kJ mol^-1

Better?

The answer I get for the question as it was written is -435 J, but the molar enthalpy change (which the question did not ask for) is -54.3 kJ mol^-1.

[Borek]

4.180 J g⁻¹ °C⁻¹ mol^-1

You overdid here

[Dan]

You overdid here

Woops! Quite right, the heat capacity was ok. Good spot Borek, 1-1 for today eh? 

[slu1986]

I'm sorry but I am still soo confused at how to set up the equation for the problem.   

[Borek]

How much heat evolved?

How many moles reacted?

Divide.