Question

Show your calculation for the molar volume of CO₂ from Trial 1-4 according to your experimental data. Show all work, value. What is the average volume of gas from all 4 trials? What is the percent error? Show your work

Trial	Pressure (atm)	Temp (K)	Reactant Mass (g)	Water Displaced (L)
1) NaHCO3	1.01	293.45	1.0	0.0004
2) NaHCO3	1.01	293.45	1.0	0.0005
3) Na2CO3	1.01	293.45	1.0	0.0005
4) Na2CO3	1.01	293.45	1.0	0.0006

Answer 1

Ans. Party A: Balanced reaction: NaHCO_3(s) + HCl_(aq) ----------> NaCl_(aq)+ H₂O_(l) + CO_2(aq)

Stoichiometry: 1 mol NaHCO₃ produces 1 mol CO₂.

Moles of NaHCO₃ = Mass / Molar mass

                                    = 1.0 g/ (84.006908g/ mol)

                                    = 0.012 mol

Now,

Number of moles of CO₂ = Number of moles of NaHCO₃        ; [see stoichiometry]

                                    = 0.012 mol

Note: Any decomposition reaction of NaHCO₃ can yield a maximum of 1 mol CO₂ per mol NaHCO₃. It’s also assumed that no CO₂ dissolves in water.

Trial 1: Volume of 0.012 mol CO₂ = 0.0004 L

So, volume of 1 mol CO₂ = Obtained volume / Moles of CO₂ evolved

                                    = 0.0004 L/ (0.012 mol)

                                    = 0.0336 L/ mol

So, molar volume = 0.0336 L

Trial 2: Volume of 0.012 mol CO₂ = 0.0005 L

So, volume of 1 mol CO₂ = Obtained volume / Moles of CO₂ evolved

                                    = 0.0005 L/ (0.012 mol)

                                    = 0.0416 L/ mol

So, molar volume = 0.0416 L

Trial 3: Volume of 0.012 mol CO₂ = 0.0005 L

So, volume of 1 mol CO₂ = Obtained volume / Moles of CO₂ evolved

                                    = 0.0005 L/ (0.012 mol)

                                    = 0.0416 L/ mol

So, molar volume = 0.0416 L

Trial 4: Volume of 0.012 mol CO₂ = 0.0006 L

So, volume of 1 mol CO₂ = Obtained volume / Moles of CO₂ evolved

                                    = 0.0005 L/ (0.012 mol)

                                    = 0.050 L/ mol

So, molar volume = 0.050 L

Average molar Volume = Average of trials 1, 2, 3 and 4

                                    = [(0.0336 + 0.0416 + 0.0416 + 0.050) / 4] L

                                    = 0.0417 L

Thus, average molar volume = 0.0417

Part B: Molar volume at given conditions-

            Pressure = 1.01 atm

            Temperature = 293.45 K

            n = 0.012 mol

Using Ideal gas equation:       PV = nRT        - equation 1

            Where, P = pressure in atm

            V = volume in L                     

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K)

1. atm x V = 0.012 atm x (0.0821 atm L mol^-1K^-1) x 293.45 K

or, V = 0.28910694 atm L / 1.01 atm = 28.62 L

Thus, molar volume under given conditions = 28.62 L

Part 3. Calculating error:

            Observed average molar volume = 0.0417 L

            Theoretical molar volume = 28.62 L

Error = Theoretical- Observed value = 28.62 L – 0.0417 L = 28.5783 L

% Error = (Error/ Theoretical value) x 100

            = (28.5783 L/ 28.62 L) x 100

            = 99.85 %