Question

When 15.0 mL of 3.00 M NaOH was mixed in a calorimeter with 13.0 mL of 3.00 M HCl, both initially at room temperature (22.0 °C), the temperature increased to 29.1 °C. The resultant salt solution had a mass of 28.00 g and a specific heat capacity of 3.74 J K-1 g-1. What is the heat capacity of the calorimeter (in J/°C)? and the molar enthalpy of neutralization per mole of HCl is -55.84 kJ mol-1

Answer 1

HCl + NaOH ==> NaCl + H2O
millimoles HCl = mL x M = 15.00 x 3.00 = 45.00 = 0.04500 moles HCl
millimiles NaOH = 13.00 x 3.00 = 39.00 = 0.03900 moles NaOH
Total volume of the solution = 15 + 13 = 28 mL amd 28 g and NaOH is the limiting regent.
Delta T = 29.1 - 22.0 = 7.1
There are three heats involved; i.e., q

Answer 2

oops. Hit the wrong button:
HCl + NaOH ==> NaCl + H2O
millimoles HCl = mL x M = 15.00 x 3.00 = 45.00 = 0.04500 moles HCl
millimiles NaOH = 13.00 x 3.00 = 39.00 = 0.03900 moles NaOH
Total volume of the solution = 15 + 13 = 28 mL and 28 g and NaOH is the limiting regent with 0.039 moles NaOH used in the neutralization.
Delta T = 29.1 - 22.0 = 7.1
There are three heats involved; i.e., q1 + q2 + q3 = 0 where
q1 = heat of neutralization = n*delta H
q2 = heat to warm the acid/base mixture = m*c*delta T
q3 = heat to warm the calorimeter = Ccal x delta T
n*dH + mc*dT + Ccal*dT
0.039*55,840 + 28*3.74*7.1 + Ccal*7.1 = 0
Solve for Ccal.