Procedure Use iron wool to polish approximately 5 cm of magnesium ribbon. Then accurately mass the sample of magnesium.
Procedure.2 With a 20 cm piece of thread, tie one end around the middle of the Mg ribbon and bend the Mg ribbon into a zig-zag pattern.
Procedure.3 Prepare a 600ml beaker two thirds full with tap water. The eudiometer tube assembly will be placed in this beaker later in the experiment.
Procedure.4 Prepare the eudiometer tube by placing 15 ml of 6M Hydrochloric acid (HCl) in the tube, then fill the tube to the top with distilled water, being careful to minimize the mixing of the acid.
Procedure.5 Quickly place the Mg ribbon in the top of the eudiometer tube and pinch the thread between the stopper and the wall of the eudiometer tube.
Procedure.6 With out letting in any air bubbles, invert the eudiometer tube assembly into the water bath; allowing the contents the ability to flow into the water bath.
Procedure.7 Make and record observations. Notice the density currents caused by the ‘sinking’ HCl. It makes the solution look oily. Once the acid reaches the Mg it begins to produce H 2 gas.
Procedure.8 Record the volume of the gas produced. 44.61 ml (read at apex of meniscus)
Procedure.9 Record the height of the water column, in millimeters, above the level of the water bath. 11.65cm high or 116.5 mm of water
Procedure.10 Record the temperature and the atmospheric pressure.
Calculations.1 The pressure of the dry H 2 gas must be calculated. There are two corrections to be made. – The water vapor mixed in with the collected H 2 must be corrected. – The pressure difference between the inside of the tube and outside the tube must be accounted for.
Calculations.2 The pressure in the tube is less than the pressure outside the tube The difference in pressure can be calculated by dividing the difference in water levels expressed in mm of H 2 O by 13.6 because mercury is 13.6 times more dense than water. 116.6 mmH 2 O 1.00 mm Hg 13.6 mmH 2 O = 8.57 mmHg
Calculation.3 The total pressure inside the tube is the atmospheric pressure minus the water column pressure difference. 774.2 mmHg – 8.57 mmHg = 765.6 mmHg
Calculations.4 Because there is H 2 and H 2 O in the tube, the pressure of the water vapor must be subtracted. Look up the water vapor pressure for the temperature of the gas. The vapor pressure of water at 22.7°C is 20.7 mmHg 765.6 mmHg - 20.7 mmHg = 744.9 mmHg
Conclusion A good conclusion includes – a summary – results – sources of error – describes an error propagation.