Question

27. 0 mL of 1.00 M HClO4 are mixed with 27.0 mL of 1.00 M LiOH in a coffee cup calorimeter. The calorimeter constant is 28.1 J/Celsius. Both were initialized at 23.31 degrees Celsius. Calculate change Hrxn (in kJ/mol) for the reaction.

Answer 1

What's the final temperature?

Answer 2

It just says, both solutions were initially 21.31 degrees Celsius, not 23.3 1 and specific heat was 4.18 J/g degrees Celsius.

Answer 3

Final temp was 27.35 degrees celsius

Answer 4

HClO4 + LiOH ==> H2O + LiClO4
mols LiOH = M x L = 0.027
mols HClO4 = M x L = 0.027

q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [(Ccal*(Tfinal-Tinitial)] where Ccal = heat capacity of calorimeter.
mass H2O = 54 mL solution = 54 grams.

dH/mol = q/0.027 and change to kJ/mol.

Answer 5

Thank You!!!