6. a Complete the following table of Trial 1 (Se Report Sheet) for determining t
ID: 591675 • Letter: 6
Question
6. a Complete the following table of Trial 1 (Se Report Sheet) for determining the molar volume of CO, and the per- cent composition of Caco, in a heterogeneous mixture. See Figure 13.4 and Equation 13.1. Record calculated values with the correct number of significant figures A. Sample Preparation and Setup Apparatus Calculation Zone 1. Mass of sample (g) 2. Mass of generator + sample before reaction ()87.719- 0226- Part C.7 C. Determination of Volume, Temperature, and Pressure of the Carbon Dioxide Gas 3. Volume of CO(g) collected (L) 4. Temperature of water ec 5. Barometric pressure (orr) 6. Vapor pressure of H,O at 20.0 C 7. Pressure of dry CO(gXtorr) -0.0377 Show calculation D. Amount of Carbon Dioxide Gas Evolved 1. Mass of generator + sample after reaction(g) 2. Mass loss of generator mass of Parr L2.3 -8HL CO, evolved (g) Show calculation. 3. Moles of CO-(g) evolved (mol Show calculation E. Molar Volume of CO, Gas Part E.2 2. Volume of CO,(g) at STP (L) Eguarion 13.4 Show calculation 3. Molar volume of CO (g) at STP (LVmol) Part E.3 Equaticn 13.2 Show calculation. F. Percent CaCO, in Mixture 1. Moles of CaCO, in sample from mol CO, generated (mol) 2. Mass of CaCO, in sample (g) Part F.2 Show calculation. 3. Mass of original sample (g) 4, Percent of Caco, in sample (%) 37.7 mL 770 tor 186 A Carbonate Analysis; Molar Volume of Carbon DioxideExplanation / Answer
From the above shown data,
C. 3. Volume (V1) = 0.0377 L
4. Temperature (T1) = 20 oC + 273 = 293 K
5. atmospheric pressure = 770 torr
6. Vapor pressure of H2O at 20 oC = 17.5 torr
7. Pressure of dry CO2 (P1) = 770 - 17.5 = 752.5 torr
= 752.5/760 = 0.99 atm
D. CO2 collected = 0.077 g
moles of CO2 = 0.077 g/32 g/mol = 0.0024 mol
E. Volume of CO2 at STP (V2) = P1V1T2/P2T2 [P2 and T2 ate pressure and temperature at STP]
= 0.99 x 0.0377 x 273.15/1 x 293 = 0.035 L
molar volume of CO2 at STP = 0.035/0.0024 = 14.58 L/mol
F. Percent CaCO3 in sample
1. moles CaCO3 = 0.0024 mol
2. mass CaCO3 = 0.0024 mol x 100.086 g/mol = 0.240 g
3. Original sample mass = 0.276 g
4. percent CaCO3 in sample = 0.240 g x 100/0.276 g = 86.96%
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