Question

A 48.71 mL volume of 1.00 M HCl was mixed with 47.19 mL of 2.00 M NaOH in a coffee cup calorimeter (with calorimeter constant = 26.1 J/°C) at 20.69 °C. The final temperature of the aqueous solution after the reaction was 27.02 °C. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following:

The total mass of aqueous solution inside the calorimeter (d_soln = 1.00 g/mL) =______g

The change in temperature (∆T) of the aqueous solution =________°C

The heat released by the neutralization reaction (this should be a positive number)=_______J

The moles of HCl reacted =_______mol

The enthalpy change (∆H) for the neutralization in kJ/mol HCl ( this should be a negative number)=_______kJ/mol HCl

Answer 1

Answer:

a)

To determine mass

Total volume = 48.71 + 47.19

= 95.9 ml

Now Mass = 95.9 ml * 1g/ml

= 95.9 grams

b)

change in temperature = T2 -T1=

= 27.02 - 20.69

= 6.33 o C

c)

Now Heat = Heat absorbed by the water + heat absorbed by the calorimeter

= 95.9 g * 4.18 K/g o C * 6.33 o C + 26.1 J/oC * 6.33 o C

= 2702.7 J

d)

HCl moles reacted = Molarity * voume of HCl

= 1.00 mol/L * 48.71 *10^-3 L

= 4.87*10^-2 mol

e)

DeltaH = Heat / HCl moles

= - 2702.7/4.87*10^-2

= - 55496.3 J/mol

= - 55.5 KJ/mol