Calorimetry Question

Question

In a calorimeter, 50.00 mL of 2.50 M NaOH was combined with 50.00 mL of 2.25 M HCl to form a solution. The temperature of the solution went from 20.0*C to 35.0*C. The heat capacity of the solution is 4.184 J/g-*C. Assume a density of 1.00 g/mL. Determine the heat of neutralization of HCl.

The last statement is what's confusing me. Do I use 50.0 grams of HCl instead of the combined 1.00*10^3 grams? Or is it actually saying to solve the heat of neutralization in kJ or J per mole of HCl?

Answer 1

It wouldn't be a combined mass of 1.00x10³ grams, but rather it would be 1.00x10² grams (50.00 ml NaOH + 50.00 ml HCl = 100 mls x 1 g / ml = 100 g). And YES, you use this mass, not 50 g. The set up would look like this:

q = mC∆T

q = (100.00 g)(4.184 J/gº)(15º) = 6276 J

Usually the heat of neutralization is given as J or kJ per mole. In this question it isn't clear what they are asking for, so I'd go ahead and report it as kJ/mol HCl

In this experiment, the HCl is limiting (there are more moles of NaOH), and the moles HCl = 0.1125 mols

∆Hneutralization = 6276 J / 0.1125 mols x 1 kJ / 1000 J = 55.79 kJ / mole