Experiment 13 Prelaboratory Assignment A Carbonate Analysis: Molar Volume of Car

Question

Experiment 13 Prelaboratory Assignment A Carbonate Analysis: Molar Volume of Carbon Dioxide Lab Soc. Nome. Desk No. Date 1. In some solid calcium carbonate samples,calcium bicarbonate, Ca(HCO,- is also present Write a balanced equation for its reaction with hydrochloric acid. 2. Experimental Procedure, Part A2a. Complete the cakulation required to appear on the Report Sheet 0here and on Report Sheet. 3 a. Experimental Procedure, Part A3b Explain how awater-filled graduated cylinder is inverted in a pan of water. b. Experimental Procedure, PartA4 What is the proper procedure for inserting a piece of glass tubing into a rubber stopper? 4. Experimental Procedure, Part Cl. How is the volume of the COCs collected (over water) measured in this 5. Athermometer is not present in the space (or volume inwhich the CO,le) is collected in this experiment. a. How then is the temperature of the CO,(g) determined in this experiment? b. The mass of a gas is elusive. How is the massofthe COlg collected over water determined inthis experiment? Experiment 13 185

Explanation / Answer

1. Balanced reaction,

Ca(HCO3)2 + 2HCl ---> CaCl2 + 2H2O + 2CO2

3. Calculations shown below

a. A glass filled cyllinder is first covered with a glass plate and then inverted in the pan.

b. Apply grease on the rubber stopper and then insert the glass piece slowly.

4. Volume of CO2 collected is measured by displacement of water in the cyllinder.

5. a. The temperature of the water is measured and then the same is used for CO2 temperature

b. The mass of CO2 which is lost after heating the same is measured as substracting final mass from initial mass.

6. Calculations,

B.

3. Volume of CO2 collected = 0.0377 L

4. Temperature = 20 oC + 273 = 293 K

5. Barometric pressure = 770 torr

6. vapor pressure H2O at 20 oC = 17.5 torr

7. Pressure of CO2 at 20 oC = 770 - 17.5 = 752.5 torr

Pressure of CO2 at 20 oC (atm) = 752.5/760 = 0.990 atm

D.

2. mass CO2 = 87.719 - 87.642 = 0.077 g

3. moles CO2 formed = 0.077 g/44 g/mol = 0.00175 moles

E.

1. Volume CO2 at STP = 0.990 x 0.0377 x 298/1 x 293 = 0.038 L

2. molar volume CO2 at STP = 0.038 L/0.00175 moles = 21.714 L/mol

F.

1. moles CaCO3 in sample = 0.00175 mol

2. mass CaCO3 in sample = 0.00175 mol x 100.1 g/mol = 0.175 g

3. mass of original sample = 0.276 g

4. percent CaCO3 in sample = 0.175 g x 100/0.276 g = 63.40%

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.