Lab Questions: Experiment 2 Lab Procedures: Take a new calorimeter from the Cont

Question

Lab Questions: Experiment 2

Lab Procedures:

Take a new calorimeter from the Containers shelf and place it on the workbench.

Take 2 M hydrochloric acid (HCl) solution from the Materials shelf and add 25.0 mL to the calorimeter.

Take a thermometer from the Instruments shelf and attach it to the calorimeter. Record the temperature inside the calorimeter in your Lab Notes.

Take 2 M ammonia (NH3) solution from the Materials shelf and add 25.0 mL to the calorimeter. Observe the temperature change in the calorimeter and record it in your Lab Notes. Make sure to record the highest temperature before the calorimeter begins to lose heat to the surroundings.

Lab Notes:

Experiment 2: Determine the Enthalpy of Neutralization of HCl by NH3

Temp of calorimeter + 25 mL 2M hydrochloric acid: 21.5 C

Temp of calorimeter + 25 mL 2M hydrochloric acid + 25 mL 2M ammonia: 33.2 C

1. What was the total mass of solution in the calorimeter after NH3 neutralizes the HCl?

50.0 g

75.0 g

25.0 g

100.0 g

21.5 °C

13.5 °C

11.7 °C

33.2 °C

3. What is the heat of solution (qsoln) for the neutralization reaction between NH3 and HCl?

1220 J

-2450 J

-1220 J

2450 J

4. What is the heat of reaction for the neutralization reaction between NH3 and HCl?

-2450 J

2450 J

2620 J

-2620 J

5. Calculate the number of moles of ammonia and hydrochloric acid. To receive full points for this question, list every step of your calculation.

6. Calculate the enthalpy of the neutralization reaction between HCl and NH3. To receive full points for this question, list every step of your calculation.

Thank you!!

Explanation / Answer

assuming the solution density =1 gm/cc, mass of the solution = density* volume= 1*(25+25)= 50 gm ( A is correct)

2. Change in temperature= final temperature-initial temperature= 33.2-21.5= 11.7deg.c ( C is correct)

3. heat of solution = mass* specific heat* temperaure difference= 50*4.184*11.7= 2248 joules (2450 joules is close answer D )

4. same as 3, 2450 Joules

5. moles of ammonia or HCl = molarity* Volume (L)= 2*25/1000= 0.05

6.Enthalpy of neutralization = 2450/0.05 Joules/mole= 49000 J/mole= 49 Kj/mole

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