Question

A 125 ml sample of 1.33M HCl and 225 ml of 0.625 M NaOH, both initially at 24.4C, are allowed to react in a calorimeter. What is the final temperature in the calorimeter? ( ΔHrxn = -57.2kJ/mol )

Answer 1

HCl + NaOH → H2O + NaCl ΔH = -57.2 kJ

(125 mL) x (1.33 M HCl) = 166.25 mmol HCl
( 225 mL) x (0.625 M NaOH) = 140.625 mmol NaOH

So NaOH is the limiting reactant.

(140.625 x 10^-3 mol NaOH) x (57.2 kJ) = 8044 J

Supposing additive volumes, and supposing the resulting solution has a specific heat near that of water, and supposing the density of the solution is near that of water:
(8044 J) / (4.186 J/g·°C) / (125 g + 225 g) = 5.49°C change

24.4°C + 5.49°C = 29.9°C final temperature