A Carbonate Analysis: otdr Volume of Carbon Dioxide Molar Volum Date Lab Sec.-_

Question

A Carbonate Analysis: otdr Volume of Carbon Dioxide Molar Volum Date Lab Sec.-_ Name A. Sample Preparation and Setup Apparatus Calculation of mass of sample for analysis (from Prelaboratory Assignment, question Desk No. _ KC10 KC03 question 2). Unknown sample no. 1. Mass of sample (g) 2. Mass of generator + sample before reaction (8) 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 fo,-collecting Trial I Trial 2 22,067322.0893 graduated cylinder (mL) 2. Final reading of volume of water in O, collecting graduated cylinder (mL) 3, Volume of#02(g) collected (L) 4. Temperature of water (C) 5. Barometric pressure (torr) 3. Vapor pressure of HO at 2 c or) 7. Pressure of dry ¢02(g) (torr) . Amount of Carbon Dioxide Gas Evolved Mass of generator + sample after reaction (g) 2. Mass loss of generators, mass go, evolved (g) . Moles of CO, evolved (mo)

Explanation / Answer

2 KClO3 ---> 2 KCl + 3 O2

Trial 1

Mass of sample :0.3120 g

Volume of O2 = 68 ml = 0.068 L

Temperature = 25C or 298 K

Pressure of barometric = 763.27 torr

Pressure of water vapor = 23.8 torr

PressurE of O2 = 763.27 torr -23.8 torr

=739.47 torr

= 0.973 atm

First calculate the moles of O2 which evolved in this reaction:

PV= nRT

n = PV/RT

= 0.973 atm* 0.068 L/ 0.08206 L –atm / mol-K*298

= 2.71*10^-3 Moles O2

Now calculate the moles KClO3:

2 KClO3 ---> 2 KCl + 3 O2

2.71*10^-3 Moles O2 * 2 mole KClO3/ 3 mole O2

= 4.065*10^-3 mole KClO3

Amount of KClO3:

Number of moles * molar mass

= 4.065*10^-3 mole KClO3 *122.55 g/mol

= 0.498 g

% of KClO3 = amount of KClO3/ sample mass *100

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