6 complete the following table of Trial I (see Report sheer) for determining the

Question

6 complete the following table of Trial I (see Report sheer) for determining the molar volume of co, per and the cent composition of Caco, in a heterogeneous mixture. See Figure 13.4 and Equation 13.I. Record calculated values with the correct number of significant figures. A. Sample Preparation and Setup Apparatus 1. Mass of sample (g) 0,216 Part C7 2. Mass of generator ple before reaction (g) -81219 C. Determination of Volume, Temperature, and Pressure of the Carbon Dioxide Gas 3. Volume of cong) collected (L) 4. Temperature of water co Part D.2 S. Barometric pressure (Morro 720 6. vapor pressure of H o at 200 7, Pressure of dry coxg)(torr) D. Amount of Carbon Dioxide Gas Evolved 1. Mass of generator sample afler reaction (s) Part D3 2. Mass loss of generator mass of C0, evolved (R) 3. Moles of CO(g) evolved (molo Part E2 E. Molar volume of Co Gas 2. Volume of coxg) at STP (L) 3. Molar volume of congo at STP (LAnolo Part E3 F. Percent Caco, in Mixture 1. Moles of Caco in sample from mol co generated (mo) 2, Mass of Caco in sample (g) Part F.2 3. Mass of original sample (g) 0.27 4. Percent of Caco, in sample (i) 377 mL 770 torr 186 A Analysis, Molor volume of Carbon Dioxide Corbonate

Explanation / Answer

c) 6) Vapor pressure of H2O at 20ºC: 17.5 torr.

7) Pressure of dry CO2(g) (torr): PCO2 = Pt - PH2O = 770 torr - 17.5 torr = 752.5 torr.

d) 2) mass of CO2 evolved (g) = 87.719 g - 87.642 g = 0.077 g

3) moles of CO2(g) evolved =

(0.0377L x 0.99atm)/(0.082L.atm/Kmol x 293.15K) =

0.037323L.atm/24.0383L.atm/mol = 1.55x10-3 moles

e) 2) Volume of CO2(g) at STP (L) = (0.0377L x 752.5 torr x 273K) / (760 torr x 293.15K) =

7744.80525 LtorrK/ 222794 torrK = 0.035 L

3) Molar volume of CO2(g) at STP = 0.035L / 1.55x10-3moles = 22.58L/mol

f) 1) moles of CaCO3 in sample from mol CO2 generated (mol) = 1.55x10-3 moles of CaCO3

2) Mass of CaCO3(g) in sample (g) = 100 g/mol x 1.55x10-3 moles = 0.155 g of CaCO3

4) Percent of CaCO3 in sample = 0.155 g / 0.276 g = 0.5616 x 100 = 56.16%

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