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Aluminum By Recycling Yiliu (Peter) Wang Teaching Assistant: Josie Bodle October 12th, 2007 Results and Discussion: Recycling Aluminum: Aluminum was recycled (in a general sense of adapting to a new use), or converted, to alum. Alum is a name for a class of compounds with the general formula of MM'(SO4)2, where M is a nonvalent cation such as K+, and M' is a trivalent cation such as Al+3. In this experiment, alum is specifically sulfate dodecahydrate [KAl(SO4)2 H2O]. In this experiment, there were intermediary chemical reactions that were needed to convert aluminum into alum. The net reaction is given by the reaction: 2Al(s) + 2KOH(aq) + 4H2SO4(aq) + 22H2O -> [2KAl(SO4)2 - 12H2O](s) + 3H2(g) In order to scale the experiment down so as to prepare a theoretical product yield of 23 grams of alum, the limiting reagent of all reactants Al, KOH, H2SO4 - was first calculated to be aluminum, which was .444 moles. Then using mole ratios through the net reaction and by conversion factors, 1.31 grams of Al was calculated to yield 23 grams of Alum. The conversion ratio of all reactants, thus, was 12 g Al (given in procedure) over 1.31 = 9.2. All reactants, then, were converted down to: .065 L 1.4 Molar KOH .026 L 9.0 Molar H2SO4 .0065 L H2SO4 aliquot (counted in the original .026 L H2SO4 solution) .0065 L H2O for rinsing residue in gravity filtration system .033 L 50/50 ethanol solution Week 1 Experiment: 1. Aluminum, from a SoBe Essential Energy Berry Pomegranate can, was cut into roughly five squares that weighed a total of 1.308 grams. It was then placed into a 600 mL beaker with 065 L 1.4 Molar KOH solution in the fume hood. Then, it was heated on a hot plate, and all the gas that was observed to rise from the chemical reaction, and not from the boiling water, was depleted after the 22:00 minute mark. Around 2/3 of the original water was lost, so 1/3, or 20.0 mL of water was put into the solution afterwards. Observations of chemical reaction on hotplate: the reaction became progressively darker and darker. Gas bubbles were emitted. There was paper residue left in the solution because it was imprinted onto the can. The balanced reaction: 2Al (s) + 2KOH (aq) + 6H2O (l) -> 2K+ (aq) + 2Al (OH)4- (aq) + 3H2 (g) The gas emitted out was then found to be hydrogen. The solution now contained paper bits, K +, Al (OH)4-, black specks of residue, and water. Using gravity filtration that was set up, the paper bits and the black impurities were then filtered out. The wrong filter paper was chosen, so the first filtration leaked bubbles and black residue into the filtrated solution. As much of the original solution was extracted from the filter paper, but a sizable amount was stuck with the paper bits and other black residue. Another gravity filtration was then done to make the black and muddy solution clear. 2. Then .026 L 9.0 Molar H2SO4 was acquired, and a .0065 L aliquot was separated form it by graduated cylinder. The aliquot was added into the filtrated solution first, and was stirred. The balanced reaction: Al(OH)4- (aq) + H+(aq) -> Al(OH)3 (s) + H2O(l) Observations: a milky white cloud was formed. After stirring, it still stayed. The milky white solid was found to be Al(OH)3 (s). 3. The remaining 9.0 Molar H2SO4 solution was slowly added in with constant stirring. The balanced reaction: Al(OH)3(s) + 3H+ (aq)-> Al+3 (aq) + 3H2O(l) Observations: after more adding solution and stirring, the milky white cloud disappeared until the solution was clear. This is because by adding more protons, it reacts with the solid to form aluminum ions again, which are soluble in water. 4. The solution was then cooled in the ice/water mixture beaker. After waiting for 10 minutes, nothing happened. Then used stirring rod to scratch the bottom of the beaker to initiate crystallization, and after another minute the solution began to crystallize. The crystallization was gradual, taking about another five minutes to fully complete. The balanced reaction: K+ (aq) + Al+3 (aq) + 2SO4-2 (aq) + 12H2O(l) -> [KAl(SO4)2 - 12H2O](s) Observation: ions Potassium gradually react with aluminum ions and sulfate ions to form hydrated crystallized solid alum. 5. The alum crystals were then filtered out of the solution by the vacuum filtration apparatus. Initials were written on the filter paper, and the solution was placed on it for filtration. Not all the solution was placed onto the filter paper because not all was emptied out of the beaker using the rubber policeman. Final balanced net reaction: 2Al(s) + 2KOH(aq) + 4H2SO4(aq) + 22H2O (l) -> [2KAl(SO4)2 - 12H2O](s) + 3H2(g) Observations: the crystallized solution was gradually reduced to fine white particles. The filter paper and the weighting boat was weighted beforehand to be 5.593 grams. 6. The filter paper, the weighting boat, and the alum particles was weighted beforehand (in week two, after drying) to be 13.500 grams. Percent Yield: (Actual/Theoretical) X 100 % [(13.500 g 5.5393 g) / 23 g] X 100% = 34% yield Conclusion: The percent yield of aluminum to alum was found to be 34%. Sources of Error: The main source of error was using the wrong filter paper in the gravity filtration, thus much of the solution stuck to the beakers and the filter paper. By filtering it again, more solution and `future alum' was lost. Secondly, the hot plate boiled away water vapor, and it carried some of the solution with it. Thirdly, every step that required the transference of chemicals left an amount of them in the beakers, thus they are lost. For example, not all the crystallized solution was transferred onto the filtration paper. Last of all, not everything reacts completely. Week 2 Experiment: Alum particles were placed into five capillaries at about .5 cm depth, and the heating device was assembled with the thermometer mounted to the capillary by a rubber band, and the hotplate heating the water bath in beaker below. The water was slowly heated in each of the five trials, but the first heating was adjusted to high on the hotplate, the second on 7, and the rest on 5. After every try, a half of the burning water was dumped out and was replaced with cold water from tap. This speeds up the trials without loss to either accuracy or precision. Table 1: Melting Point of Alum Crystals Trial Number 1 2 3 4 5 Hot Plate Setting On High Notch 7 Notch 5 Notch 5 Notch 5 Melting Point of Alum 90.0o C 91.0o C 91.5o C 91.4o C 91.7o C The trend was if the hot plate setting was higher, the more inaccurate, imprecise, and lower the melting point of alum would appear to be. Thus, in order to calculate precision and accuracy, trial #1 and #2 were not included given that they were on higher and different settings. Only notch 5 setting on hot plate was used for calculations because only with the same settings (or same independent variables) can precision and accuracy be even used. Precision Calculations: Precision was calculated using the formula for relative average deviation. Average of the three trials: 91.5o C + 91.4o C + 91.7o C = 91.5o C |91.5o C - 91.5o C | + |91.4o C - 91.5o C | + |91.7o C - 91.5o C | / (3) = .100 o C (.100 o C / 91.5o C) X 100% = .109% relative average deviation. The results, then, were determined to be adequately precise. Accuracy: Only after matching the calculated melting point with the actual melting point of alum can the calculation of accuracy be made possible. The actual melting point was found to be 92.5 o C Percent error: |(91.5o C - 92.5 o C) / 92.5 o C | X 100% = 1.08% The results, then, were determined to be adequately accurate. Conclusion: The experimental results were adequately precise, and they were relatively close, or accurate, compared to the actual melting point. Thus, they are relatively pure given the small percentage error between the mean and actual melting point. Sources of Error: Possible sources of error include incomplete gravity filtration of residues, vacuum filtration that filter in other particles, and impurities and other chemicals that have accumulated during the week. The rest of the alum was then put into vials. Measured vial: 16.465 g Measured vial and alum: 23.459 g ... View Full Document

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