# Preparation of Alum from Aluminum Metal Essay example

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LABORATORY REPORT 4 PREPARATION OF ALUM FROM ALUMINUM METAL Huy Nguyen October 2nd, 2012 The objective of the laboratory is to synthesize alum (KAl(SO4)2.xH2O) from aluminum powder and to determine the proportion of water in the alum crystals. Alum is a product from the reaction between potassium hydroxide and sulfuric acid. The reaction include several steps, as followed: Aluminum powder reacts with potassium hydroxide to generate Al(OH)4- ions and release hydrogen. 2 Al(s) + 2 KOH(aq) + 6 H2O 2 K[Al(OH)4](aq) + 3 H2 (g) A gelatinous precipitate of aluminum hydroxide was created when sulfuric acid was added to the aqueous solution of Al(OH)4- ions. 2 K[Al(OH)4](aq) + H2SO4 (aq) 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2O…show more content…
The crystals were heated at maximum heat for 5 minutes. The crucibles were placed back to the desiccator. After cooling to room temperature, the masses of the contents inside the crucibles were carefully weighed. Results The masses of alum, KAl(SO4)2 and water recorded were given in Table I. Table I. Masses of Alum, KAl(SO4)2 and water in two different crucibles. | Crucible 1 | Crucible 2 | Alum | 0.5000 g | 0.5000 g | KAl(SO4)2 | 0.2721 g | 0.2696 g | H2O | 0.2279 g | 0.2304 g | x= nwaterndry product | 12.00 | 12.24 | According to the values of x obtained from the table above, the average result of x is 12.12. We can define the formula of alum as KAl(SO4)2.12,12H2O (Molar Mass M = 476.16 gmol-1). Finding the formula of alum makes it possible to calculate the theoretical yield and the percent yield of alum. After calculations from the equations demonstrated in the introduction, the theoretical number of moles of alum would be 0.019 moles. The theoretical yield, as a result, would be mtheoretical = 9.69 g. The actual yield recorded after the laboratory was 4.77 g. Combining all the yields gives us the final result of the percent yield: 52,71%. Discussion Several steps of heating the alum crystals and calculations took place to find out the formula of alum. Concerning the first crucible, an amount of 0.5 g of alum was added to the crucible. After