I need help with Page 14 Discussion questions a-g ASSIGNMENT 5 (Gas 01) The Dete

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I need help with Page 14 Discussion questions a-g ASSIGNMENT 5 (Gas 01) The Determination of the Volume of a Mole of Oxygen MATERIALS Rubber stopperitubing assembly to fit 500 mL flask; large graduated cylinder and thermometer from stockroom, as needed unknown KCI03 mixture (bring a test tube) BE SURE TO WEAR PROTECTIVE GOGGLES 1. Avogadro's Principle tells us that a mole of a gas will occupy the same volume as a mole of any other gas, if the conditions are the same. In this assignment you will measure the mass and volume of oxygen produced by the decomposition of KCIO3 mixed with inert substances. The decomposition proceeds more rapidly and safely when catalytic amounts of manganese dioxide are present. Molten KCIO3 decomposes with the loss of oxygen and the mass lost is a measure of the oxygen evolved. The other product is potassium chloride, KCI. You will use the mass of oxypen to caliculate the percent KCIOs in the mixture and then determine the volume occupied by a mole of oxygen at standard conditions. EXPERIMENT Obtain your unknown from the stockroom in a clean, dry test tube. Set up your apparatus as you are instructed. See diagram on page 12 and ask the instructor for approval before you begin to heat anything. Be sure to have well-fitting stoppers; the apparatus must be gas-tight. You will have been given instruction in the safe handling of KCIO3. The connection between the test tube and the flask should be flush with the botiom of both stoppers; the delivery tube should reach nearly to the bottom of the flask. If you have to shift the tubing in the stoppers, begin by rotating the stopper, then gently push or pull on the tubing while rotating the stopper. It is advisable to protect your hands by wrapping the tubing in paper towel. Be certain that any test tube you heat is Pyrex or Kimax ("hard glass") and that the mouth of the test tube points at least slightly "up" chlorates react vigorously with rubber and fire is a poor beginning for an experiment Weigh a clean, dry Pyrex test tube. Put about half of your unknown in the test tube and weigh it again Assemble the apparatus as you are told, being sure to leave the test tube off until you are ready to adjust the pressure in the system. Fill the flask to within 3 cm of the bottom of the stopper and place a 600 mL beaker full of water where the delivery tube can reach nearly to the botiom. Establish a siphon as descibed by your instructor, then raise the beaker to restore the level of water in the fiask. Clamp the rubber connector to prevent siphoning and connect the test tube. Be sure all connections are tight and open the pinch-clamp to allow free flow. Raise the beaker untl the level of water in the beaker is the same as the level inside the flask (equalize the water levels) and close the pinch clamp. The pressure is now the same inside the flask as in the atmosphere outside. Pour the water out of the 600 mL beaker and shake it once. Replace the delivery tube in the beaker. You are now ready to begin the experiment First RELEASE THE CLAMP. A small amount of water should run out into the beaker but not more than one or two mL. If it continues to flow (or drip), you must reassemble the system and equalize the pressures before you begin.] Holding the burner in your hand so as to be able to control the heating rate, heat no more than the lower two- thirds of the test tube until you see some melting of the mixture occur. NEVER leave the burner sitting under the test tube. Decrease the heating rate a little but do not stop heating until enough water has been delivered to cover the end of the delivery tube. Continue to heat moderately until no more water flows into the beaker. Stop heating and allow the system to coo Leave the apparatus unclamped untl the mixture has cooled to room temperature. Once again equalize the levels of water-this time you may have to raise the flask-and clamp the delivery tube. Now you may disassemble the system in order to weigh the test tube and its contents, measure the volume of water in the beaker, and measure the emperature of the water in the flask. (You measure the temperature of the water in the flask because it was in contact with the warm gas.) Record the data in your notebook. The test tube and contents are heated again (t is not necessary to collect the oxygen) to check for complete reaction (constant mass) Clean and dry another Pyrex test tube. Weigh it put the rest of the sample in i, and weigh it again. Reset the system as above and repeat the entire experiment a second time. 12

Explanation / Answer

a) Pressure inside flask = (H2-H1)* d * g

Where H2-H1 is the height difference in two flasks. d = density, g = acceleration due to gravity

b) The flask is initially at atmospheric pressure (760 torr) and initial temperature is 31 0 C. If the flask is connected to a vacuum pump which maintains a constant pressure of 25.0 torr, then the pressure inside flask is reduced from 760 torr to 25 torr. The total volume of the gas inside flask remains fixed. Under constant volume, pressure and temperature (in kelvin scale) are proportional to each other. As pressure decreases, temperature should increase.

The relation between temperature (in K scale) and pressure is :

P1*T2 = P2*T1

P1 = initial pressure = 760 torr, T1 = initial temperature = 31 0C = (31+273) K = 304 K

P2 = final pressure = 25 torr, T2 = final temperature = ?

760 torr * T2 = 25 torr * 304 K,    So, T2 = (25*304)/ 760 K = 10 K

So, temperature inside flask is 10 K = (10-273) 0C = -263 0C

c.

The balanced equation is : 2NO + O2 =2NO2

i) If 2 moles of NO reacts then 1 mole of O2 is consumed and 2 moles of NO2 is produced

So, for 1 mole of reacting NO: 0.5 mole of O2 is consumed and 1 mole of NO2 is produced

ii) At STP, 1 mole of any ideal gas occupies a volume of 22.4 litres. Assuming NO< O2 and NO2 are all ideal gases,

1 mole of NO has volume of 22.4 L, 0.5 mole of O2 has volume of 22.4*0.5 Litres= 11.2 Litres, 1 mole NO2 has volume of 22.4 Litres

iii) 1 mole NO = 1 mole * 30 (g/mole) = 30 grams NO

0.5 mole O2 = 0.5 mole *32 (g/ mole) = 16 grams O2

1 mole NO2 = 1 mole* 46 (g/ mole) = 46 grams NO2

d) Molar mass of HCl = 36.5 grams/ mole

At STP. 22.4 litres of HCl = 1 mole of HCl

At STP, 4.55 litres of HCl = (4.55/ 22.4) moles of HCl = 0.203 mole of HCl = 0.203 mole * 36.5 (g/ mole) = 7.41 grams

e) Density of the gas at STP is 0.00481 g/ mL

So, at STP 1 ml of the gas has the mass of 0.00481 grams

At STP 22.4 Litres (or 22400 mL) of the gas 22400 * 0.00481 grams = 107.744 grams

And at STP 22.4 litres of an ideal gas conrains 1 mole

So, 1 mole of the gas has the mass 107.744 grams. The molecular weight of the gas is 107.744 g/ mole

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