Question

When 50.08 g of 1.00 M NaOH at 23 C is added to 50.22 g of 1.00 M HCl at 23 C, the resulting temperature of the mixture is 29.7 C. (Assume that the calorimeter constant is equal to zero. Assume that the solutions have the same specific heat capacity (4.184 J/g·K) and density (1.00 g/mL) as water. Take care with signs.

a) What is ΔT for this process? b) Calculate qsolution (in J) for this process. c) Calculate qreaction (in J) for this process. d) Calculate the moles of NaOH and the moles of HCl used in the reaction. Determine whether NaOH or HCl is the limiting reactant. e) What is the enthalpy change (ΔHrxn, in kJ/mol) for this process per mol of the limiting reactant? f) Is the reaction of NaOH with HCl an endothermic or exothermic process

Answer 1

Step-by-step explanation:

a) ΔT = final temperature - initial temperature = 29.7°C - 23°C = 6.7°C.

b) qsolution = mass of solution x specific heat capacity x ΔT = (50.08 g + 50.22 g) x 4.184 J/g°C x 6.7°C = 574.47 J.

c) qreaction = - qsolution = -574.47 J.

d) moles of NaOH = mass of NaOH / molar mass of NaOH = 50.08 g / (23 g/mol) = 2.17 mol

moles of HCl = mass of HCl / molar mass of HCl = 50.22 g / (36.46 g/mol) = 1.37 mol

Since the moles of NaOH is greater than the moles of HCl, HCl is the limiting reactant.

e) ΔHrxn = qreaction / moles of limiting reactant = -574.47 J / 1.37 mol = -418.24 kJ/mol.

f) ΔHrxn is negative, so the reaction of NaOH with HCl is an exothermic process.