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Unformatted text preview: Results and Discussion: A given procedure was scaled down in order to prepare a theoretical yield of 23 g of alum. In order to synthesize the alum from aluminum, both gravity and vacuum filtrations were performed. After being crystallized and isolated, the percent yield and melting point of the solid alum were determined. An aluminum soda can was cut into tiny strips. The mass of these strips was found to be 1.295 grams using a highly calibrated balance. After finding the mass, the strips were then placed into an empty 600-mL beaker. In the fume hood, 67 mL of 1.4 M KOH was added to the 600-mL beaker containing the aluminum strips and the mixture was heated gently for about 30 minutes. A reaction occurred while heating, causing the aluminum strips to dissolve and a gas (H 2 ) to be produced. This reaction is demonstrated by the following equation: 2 Al (s) + 2 KOH (aq) + 6 H 2 O (l) 2 K → + (aq) + 2 Al(OH) 4- (aq) + 3 H 2(g) About 40% of the evaporated water was replaced with distilled water. Then, the product of the previous reaction was filtered using gravity filtration, and a faint clearish-yellow solution was obtained in a clean beaker. 26.5 mL of 9M H 2 SO 4 was measured using a graduated cylinder and was gradually added to the clearish-yellow solution in the beaker while it was being warmed on a hot plate. After the first 6.5 mL of sulfuric acid was added to the beaker, a thick, white precipitate formed, which can best be understood in terms of the following equation.which can best be understood in terms of the following equation....
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This note was uploaded on 02/27/2008 for the course CHEM 2070 taught by Professor Chirik,p during the Fall '05 term at Cornell University (Engineering School).
- Fall '05