Preparation of Alum from Aluminum Metal Essay

The objective of the laboratory is to synthesize grad (KAl(SO4)2.x weewee) from atomic number 13 powder and to determine the proportion of piddle system in the ammonia grade crystals. alumna is a product from the reaction between kilobyte hydroxide and sulfuric harsh. The reaction include several steps, as followedAluminum powder reacts with special K hydroxide to generate Al(OH)4- ions and release hydrogen. 2 Al(s) + 2 KOH(aq) + 6 H2O 2 KAl(OH)4(aq) + 3 H2 (g)A gelatinous strike of atomic number 13 hydroxide was created when sulfuric biting was added to the aqueous solution of Al(OH)4- ions. 2 KAl(OH)4(aq) + H2SO4 (aq) 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2OLater, excessive addition of the acid causes the precipitate to dissolve in the solution. 2Al(OH)3 (s) + H2SO4 (aq) Al2(SO4)3 (aq) + 6 H2OPrecipitation of alum was resulted from change in nut piddle bath. K2SO4 + Al2(SO4)3 + 2x H2O 2 KAl(SO4)2.xH2OIt is noticeable that alum is a provide (a hydrate consists of pis s molecules in its ionic structure), which leads to its solubility in piss supply. However, a minimum amount of cold water will cause the alum to crystallize. The amount of water incorporated in the alum structure should be clearly defined to derive the full formula of alum, which makes it realistic for calculations of suppositional, real and percent wages of alum.Experimental MethodsThe experiment was constructed based on the guidelines from Franklin and marshall Lab Manual1. In a 400 mL-beaker, 0.5 g of aluminum and 2.01g of potassium hydroxide was prepared and mixed together. An amount of 25 mL of distilled water was poured into the beaker in the hood. The mixture was then continuously stirred to help mobilise the heat generated from the exothermic reaction. As observed, hydrogen was liberated from the solution, along with aluminum powder gradually darkening and disintegrating into insoluble flakes. It took the solutions 15 minutes to perpetrate when there were no signs of hydrogen released. The solution was then filtered into a unseasoned 250 mL beaker. The resi out-of-pocket left on the filter report was conservatively washed into the filtrate.A portion of 10 mL of 9M sulfuric acid was added slowly and attentively to the filtrate, with low stirring. The presence of acid will antagonise the solution, generating a gelatinous precipitate known as Al(OH)3. The precipitate was by and by dissolved when excessive addition of acid was poured into the solution, combined with gentle hotness on hot plate. The acidity of the solution was confirmed when tested with litmus paper the paper turned into red. The solution was filtered for the act time to disdain any undissolved residues remaining.The solution was coif aside to cool at room temperature. The crystallization process was conducted by placing the solution beaker into an ice water bath for 20 minutes. later on crystallization, white, soft crystals were formed. The mixture was filtered throu gh a Buchner funnel.A wash solution was prepared by combining 5 mL of ethyl alcohol and 5 mL of distilled water. The crystals were washed double with proper wash solution. Then, the solution was put through suction once more to dry out completely. The crystals were spread in a recrystallization disk. Large crystals were busted into small ones with a stapula. The crytals were allowed to air dry in one week.The lading of the air-dried crystals was then recorded. Two porcelain crucibles were supported on ceramic triangles and alter to red heat with a Bunsen burner for 10 minutes each. The crucibles were setaside cool, then was placed into the desiccator to cool to room temperature. Their weighs were recorded. An amount of 0.5 g of the crystallized alum was placed into each of the crucibles. The crucibles (with alum inside) were care amply heat on ceramic triangles to red heat. The alum inside the crucibles appeared to melt, transforming into a mannequin of liquid solution. After 5 to 10 minutes of continuous and gentle heating, the content inside the crucibles started to solidify again, yielding white, soft crystals. The crystals were heated at maximum heat for 5 minutes. The crucibles were placed back to the desiccator. After cooling to room temperature, the pottyes of the contents inside the crucibles were carefully weighed.ResultsThe 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 2Alum 0.5000 g 0.5000 gKAl(SO4)2 0.2721 g 0.2696 gH2O 0.2279 g 0.2304 gx= nwaterndry product 12.00 12.24According to the quantifys of x obtained from the turn off 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 conjectural yield and the percent yield of alum. After calculations from the equations demonstrate in the introd uction, 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 aft(prenominal) the laboratory was 4.77 g. Combining all the yields gives us the final result of the percent yield 52,71%.DiscussionSeveral steps of heating the alum crystals and calculations took place to commence out the formula of alum. Concerning the first crucible, an amount of 0.5 g of alum was added to the crucible. After heating, there was 0.2521 g of contents (KAl(SO4)2) left in the crucible. That means there was 0.2479 g of H2O fully evaporating. In this case, x= nH2Ondry product= 0.2279180.2721258= 12.00. Concerning the second crucible, an amount of 0.5 g of alum was added to the crucible. After heating, there was 0.2496 g of contents (KAl(SO4)2) left in the crucible. That means there was 0.2504 g of H2O fully evaporating. In this case, x= nH2Ondry product= 0.2304180.2696258= 12.24.The average result of x x= 12.00+ 12.242= 12.12. With calculations concerning the masses of contents in the crucibles before and after heating, it is observed that 12.12 molecules of water in a mole of alum. The general formula of alum, therefore, is KAl(SO4)2.12.12H2O. The literature value of portions of water molecules in alum is 12, which makes the formula of alum KAl(SO4)2.12H2O. The proximity of the work out result and the literature result reflected to efficiency and accuracy of the laboratory.Through a series of chemical reactions, alum (the double salt with incorporated water molecules, with the calculated formula of KAl(SO4)2.12H2O) was formed from aluminum powder, potassium hydroxide and sulfuric acid. The reactions lead to the formation of alum are summarised as followed (I) 2 Al(s) + 2 KOH(aq) + 6 H2O 2 KAl(OH)4(aq) + 3 H2 (g) (II) 2 KAl(OH)4(aq) + H2SO4 (aq) 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2O (III)2Al(OH)3 (s) + H2SO4 (aq) Al2(SO4)3 (aq) + 6 H2O(IV)K2SO4 + Al2(SO4)3 + 24 H2O 2 KAl(SO4)2.12H2OThe theoretical yield was accumulated over a few steps in that location are 0.019 moles in 0.5 g of Aluminum. Similarly, there are 0.036 moles in 2.01 g of potassium hydroxide. We used a portion of 10 mL of 9M sulfuric acid, meaning that we use 0.09 moles of sulfuric acid.In reaction (I) that potassium reacted with aluminum powder with the presence of water, the aluminum played the role of the limiting reagent. In reaction (II) that sulfuric acid was added into the solution of Al(OH)4- ions, the ions were the limiting reagents. The gelatinous precipitate formed in reaction (II) by pouring in acid was soon dissolved in the solution in the reaction (III) by the addition of excessive sulfuric acid. The alum crystals were formed in the reaction (IV) by cooling. From the four reactions, we can easily see that the number of moles of alum formed is equal to the number of moles of aluminum in the aluminum powder. nalum = naluminum = 0.019 moles.The theoretical yield is the product of the numbe r of moles and alums molar mass malum= n M= 0.019 476.16= 9.05 (g). The actual yield is 4.77 g (as stated in the results). The percent yield is calculated by dividing the actual yield by the theoretical yield %Yield= Actual YieldTheoretical Yield = 4.77 g9.05 g = 52.71%.well-nigh 47% of alum was lost during the crystallization. From 0.5 g of aluminum, 2.01 g of potassium hydroxide and 10 mL of 9M sulfuric acid at the beginning, the product obtained after crystallization was only 4.77 g of alum, compared to the theoretical value of 9.05 g. A real amount of alum was lost during filtration, suction and crystallization, because of the fact that the filter paper was not wet enough and the crucibles were not dry enough due to short maximum heating time.References1. Franklin and Marshall College Chemistry 111/112 Laboratory Manual, choke 2012/Spring 2013, p. 39-41.