You place 40.6 ml of 0.659 m naoh in a coffee- cup calorimeter at 25.00°c and add 71.3 ml of 0.659 m hcl, also at 25.00°c. after stirring, the final temperature is 28.57°c. [assume the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 j/gk)]. calculate the change in enthalpy (latex: \deltaδh) of the reaction in kj/mol of water formed. enter to 1 decimal place.

ΔH of reaction is 62,3 kJ/mol

Explanation:

For the reaction:

HCl + NaOH → H₂O + NaCl

The heat transfer in the reaction (q) =

q = Specific heat × mass × ΔT

The mass is 40,6 mL + 71,3 mL × (1g/mL) = 111,9 g

And ΔT is 28,57 - 25,00 = 3,57°C

q = 4,184 J/gK × 111,9 g × 3,57°C = 1671 J

The moles of NaOH are:

0,0406 L×0,659 M = 0,0268 moles

And the moles of HCl are:

0,0713 L ×0,659 M = 0,0470 moles

The moles of reaction are the moles of NaOH because are the limiting reactant. 0,0268 moles

ΔH of reaction is q/moles of reaction. Thus:

ΔH of reaction =  1671J/0,0268 moles = 62350 J/mol ≡ 62,3 kJ/mol

I hope it helps!

the answer is 1 i believe