leats of Formation an* Ine Experiment 17 Molar Volume of Oxygen Background: The
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leats of Formation an* Ine Experiment 17 Molar Volume of Oxygen Background: The relationship of pressure, volume and temperature for a gas sample has been extensively studied When the number of moles of gas is taken into consideration as well, these relationships can be expressed in the ideal gas law: PV n RT At STP (1.0 atm and 0'C) this constant is 22.4 l/moke. In this experiment we will determine the molar volume of oxygen gas (02) at room temperature. From the molar volume you can determine the density oro2gasatroomtemperature . motor vo lemtdins The reaction that you will be using to generate oxygen is the decomposition with heat and MnO catalysis of potassium chlorate. 2KCio, (s) MnO2 (s, catalyst 2KCl (s) + 3 02u) . You will also determine the percentage of potassium c our We will measure the volume of wet oxygen (collected over water) by measuring the volume of water that the oxygen displaces. Atmospheric pressure, which will be equal to the wet oxygen pressure, will be measured using a barometer. The water temperature (in thermal equilibrium with the gas) will be taken as the gas temperature. The mass of oxygen will be determined from PV=nRT- Safety Tips, please read before going to the next page, which are the procedures for the lab. DO NOT MELT THE RUBBER ON THE TEST TUBE CLAMP DO NOT LEAVE FLAME UNATTENDED WEAR GOGGLES WHILE IN THE LABORATORY! Potassium chlorate is a powerful oxidizing agent and reacts readily with easily oxidized substances such as lubricating oil, rubber or glycerin. Careless handling of it could conceivably result in an explosion. Therefore, in performing this experiment, take the following precautions: 1. Use a Pyrex test tube that is clean and dry. To clean the test tube, wash with soap and water, rinse with tap water, rinse three or four times with distilled water, then flame dry it. DO NOT wipe the inside with a paper towel or blow compressed air into it. Compressed air will introduce small oil droplets with possibly disastrous results 2. Use a PAPER FUNNEL to introduce the potassium chlorate mixture into the tube, so that none is deposited high up on the wall where it might come in contact with the rubber stopper. 3. Clamp the test tube so that it tilts UPWARD at about a 30° angle from the horizontal.Explanation / Answer
11) Partial pressure of water at 294.15 K (= 294.15 K – 273.15 K = 21.0 K) = 18.6 torr = (18.6 torr)*(1 atm/760 torr) = 0.02447 atm (1 atm = 760 torr).
12) Pressure of oxygen at 294.15 K = (atmospheric pressure) – (partial pressure of water at 294.15 K) = 0.97553 atm.
13) Moles of oxygen gas:
Use the ideal gas law: P*V = n*R*T where P = 0.97553 atm; V = 0.12 L; T = 294.15 K.
n = P*V/RT = (0.97553 atm)*(0.12 L)/(0.082 L-atm/mol.K).(294.15 K) = 0.0048533 mole.
14) Molar volume of oxygen = (volume of oxygen gas)/(moles of oxygen gas) = (0.12 L)/(0.0048533 mole) = 24.7254 L/mol.
15) Density of oxygen gas = (mass of oxygen gas released)/(volume of oxygen gas) = [(mass of test tube with unknown and MnO2) – (mass of test tube with unknown and MnO2 after heating)]/(volume of oxygen gas) = (45.457 – 45.393)g/(0.12 L) = (0.064 g)/(0.12 L) = 0.5333 g/L.
16) Grams of KClO3 in the mixture:
Consider the balanced stoichiometric equation for the reaction.
2 KClO3 (s) --------> 2 KCl (s) + 3 O2 (g)
As per the stoichiometric equation,
2 mole KClO3 = 3 mole O2.
Therefore, 0.0048533 mole O2 = (0.0048533 mole O2)*(2 mole KClO3/3 mole O2) = 0.0032355 mole KClO3.
Molar mass of KClO3 = 122.55 g/mol.
Mass of KClO3 in the mixture = (0.0032355 mole)*(122.55 g/mol) = 0.396 g.
17) Percent KClO3 in the unknown = (mass of KClO3 in the mixture)/(mass of the mixture)*100 = (0.396 g)/(2.512 g)*100 = 15.764%
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