In this exercise, weighed samples of a solid unknown containing NaCl and NaHCO 3

Question

In this exercise, weighed samples of a solid unknown containing NaCl and NaHCO3react with hydrochloric acid. The volume of the CO2 liberated by the reaction in the gas phase is measured with a gas collection syringe. From the volume we can calculate the number of moles of CO2 in the gas phase using the ideal gas approximation.

Since the CO2 is generated in an aqueous environment (aqueous HCl), some CO2will dissolve in the liquid phase. The amount of CO2 dissolved in the liquid phase is computed using Henry's Law which requires knowing the partial pressure of CO2. As described in the exercise, this requires knowing the entire system gas volume in addition to the initial and final syringe readings.   

The entire system gas volume is the sum of the quantities labeled Vtube and Vsyr in the diagram below, corrected for the liquid volume (the aqueous HCl).

The table below contains the data collected in one run of the reaction between an unknown and hydrochloric acid:

Initial Weight of container and unknown

Torr

The following questions deal with the above data. The appropriate units are specified in the questions. Pay close attention to the number of significant figures in your answers.

1). What is the volume of CO2 captured in the gas phase (in mL)?

2).What is the number of mmols of CO2 captured in the gas phase? (Remember to account for the vapor pressure of water.)

3). What is the weight of the sample in grams?

Value Units Gas Constant 0.0821 L-atm/mol K Absolute Zero (0 K) -273.15 oC Atmosphere 760.0 Torr Molar Mass of NaHCO3 84.01 g/mol Henry's Law Constant for CO2in water 3.2 X 10-2 mol/L-atm

Explanation / Answer

1) volume of CO2 captured in the gas phase (in mL)=Final volume of syringe-Initial volume of syringe=50.7-5.0=45.7ml

2)moles of CO2 captured in the gas phase=n=PV/RT   ( SINCE PV=nRT) ,

P=Pressure=764.4 torr=764.4 * 0.001315 atm=1.0051 atm, temperature=26.5 +273=299.5K

                    n=1.0051 atm *45.7ml/0.082 L atm/Kmol*299.5K=1.87 moles

3)partial pressure of CO2=pCO2

pCO2/ptotal=Vgas/Vtotal

pCO2=47.5/92.3 * 1.0051 atm=0.517 atm

using henry's law conc of CO2 soluble in HCl(10 ml)=C

C= k* pCO2=3.2 * 10-2 mol/L atm * 0.517 atm=0.00165 mol /L

so moles of CO2=.0.00165 *10=0.0165 mol(absorbed in HCL)=0.165mol * 44g/mol=72.8 g

mass of CO2 in gas phase=1.87 moles * 44g/mol=85.28g

total weight of CO2=72.8+85.28=155.08 g

3)mass of sample =18.6185g-18.4395g=1790 g (from table)

                                                                

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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