Question

Part A The aluminum cup inside your calorimeter weighs 42.95 g. You add 60.51 g of 1.0 M acetic acid solution and 49.33 g of 1.0 M sodium hydroxide solution to the calorimeter. Both solutions have an initial temperature of 19.9 oC, and the final temperature after addition is 26.7 oC. What is the heat of neutralization for the amounts of reactants used, in units of J? Assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup. the density of the two solutions is the same as that of water: 1.00 g/mL. the heat capacity of the two slutions is the same as that of water: 4.184 J g-1 oC-1. Perform all calculations without rounding, but then provide your answer to the correct number of significant figures. Part B What are the moles of the limiting reagent? Part C What is the molar enthalpy of neutralization determined from the calorimeter experiment, in units of kJ/mol? Use unrounded values in your calculation, but then provide your answer to the correct number of significant figures.

Answer 1

Part A : heat of neutralization = -3130 J

Part B : moles of the limiting reagent = 0.049 mol

Part C : molar enthalpy of neutralization = -63 kJ

Explanation

Part A

Total mass of solution = (mass acetic acid) + (mass sodium hydroxide)

Total mass of solution = (60.51 g) + (49.33 g)

Total mass of solution = 109.84 g

Heat gained by solution = (Total mass of solution) * (specific heat of solution) * (final temperature - initial temperature)

Heat gained by solution = (109.84 g) * (4.184 J/g.^oC) * (26.7 ^oC - 19.9 ^oC)

Heat gained by solution = 3125.08 J

Heat of neutralization = -(Heat gained by solution)

Heat of neutralization = -(3125.08 J)

Heat of neutralization = -3125.08 J