Unknown sample no. Trial 2 Trial 1. Mass of sample (g) 219- 204 2. Mass of gener

Question

Unknown sample no. Trial 2 Trial 1. Mass of sample (g) 219- 204 2. Mass of generator sample before reaction (g) 3. Instructor's approval of apparatus C. Determination of volume, Temperature, and Pressure of the Carbon Dioxide Gas 1. Initial reading of volume of water in con-collecting graduated cylinder (ml) 2. Final reading of volume of water in co, collecting graduated cylinder (mL) 49 mi 3. Volume of Co20g collected (LI 2-1 4. Temperature of water 5. Barometric pressure (torr) 6. Vapor pressure of H20 at °C (torr) 7. Pressure of dry CO20g) forr D. Amount of Carbon Dioxide Gas Evolved 1. Mass of generator sample after reaction (g) 2- 2. Mass loss of generator mass CO2 evolved (g) 3. Moles of Con evolved (mo) maes mol MM Experiment 13 187

Explanation / Answer

PART E)

Volume of CO2 at STP

mol of CO2 --> 0.0038628 (trial1) and 0.023856 (trial 2)

at STP:

by definition 1 mol of any gas = 22.4 L

so

trial (1) 0.0038628 mol of CO2 --> 0.0038628 *22.4 L = 0.08652672 literst = 86.56 mL

trial (2) 0.023856 mol of CO2 --> 0.023856 *22.4 L = 0.53437 literst = 534.6 mL

3.

Molar Volume of CO2 at STP -->

by definition

V molar = V total / mol sample = 534.6 /0.023856 = 22400 mL /mol

4.

Averate molar Volume =  22400 mL /mol

F)

1 mol of CO2 --> 1 mol of CaCO3

so

trial 1 --> 0.0038628 mol of CO2 --> 0.0038628 mol of CaCO3

trial 2 --> 0.023856 mol of CO2 --> 0.023856 mol of CaCO3

2

MW of CACO3 = 100.0869

mass of CaCO3 = mol*MW = 100.0869 *0.0038628 = 0.386 g of CACO3

mass of CaCO3 = mol*MW = 100.0869 *0.023856 = 2.38 g of CaCO3

Mass of sample -->

0.21 and 0.20 g

% CaCO3 = 0.386 /0.21*100 = 183%

% CaCO3 = 2.38 /0.20*100 = 1190 %

AVerage CaCO3 = 686.8 %

Clearly, there was an error in any measurement/calculation (done before)

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