Preparation of Alum from Aluminum Metal Essay

The mark of the laboratory is to synthesize ammonium alum (KAl(SO4)2.x pee) from aluminium powder and to determine the proportion of water in the alum crystals. Alum is a product from the chemical reception between potassium hydroxide and sulphuric savage. The reaction take on several go, as followedAluminum powder reacts with potassium hydroxide to generate Al(OH)4- ions and release enthalpy. 2 Al(s) + 2 KOH(aq) + 6 H2O 2 KAl(OH)4(aq) + 3 H2 (g)A gelatinous precipitate of aluminium hydroxide was created when sulfuric dit was added to the aqueous resolution 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 amazes the precipitate to dissolve in the solution. 2Al(OH)3 (s) + H2SO4 (aq) Al2(SO4)3 (aq) + 6 H2OPrecipitation of alum was coreed from cooling in starter water bath. K2SO4 + Al2(SO4)3 + 2x H2O 2 KAl(SO4)2.xH2OIt is noticeable that alum is a hydrate (a hydrate consists of water molecules in its ioni c structure), which leads to its solubility in water. 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 delimitate to derive the full formula of alum, which makes it possible for calculations of theoretical, material and percent yield of alum.Experimental MethodsThe experiment was constructed found 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 prepa going and mixed together. An amount of 25 mL of distilled water was poured into the beaker in the hood. The mixture was then interminably stirred to help disperse the kindle 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 complete when there were no signs of hydrogen released. The solution was then filtered into a new 250 mL beaker. The residue leftfield on the filter constitution was guardedly damped into the filtrate.A portion of 10 mL of 9M sulfuric acid was added slowly and attentively to the filtrate, with gentle stirring. The presence of acid will neutralize the solution, generating a gelatinous precipitate known as Al(OH)3. The precipitate was later dissolved when excessive addition of acid was poured into the solution, combined with gentle heating 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 split second time to eliminate any undissolved residues remaining.The solution was set 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 crystallization, white, soft crystals were formed. The mixture was filtered through a Buchner funnel.A wash solution was prepared by unite 5 mL of ethyl alcohol and 5 mL of distilled water. The crystals were washed twice with proper wash solution. Then, the solution was put through suction again to dry divulge completely. The crystals were spread in a recrystallization disk. Large crystals were broken into small ones with a stapula. The crytals were allowed to air dry in one week.The weight of the air-dried crystals was then recorded. Two porcelain crucibles were supported on ceramic triangles and heated to red heat with a bunsen burner 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 carefully heated on ceramic triangles to red heat. The alum inside the crucibles appeared to melt, transforming into a kind of liquid solution. later on 5 to 10 minutes of free burning and gen tle 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 masses 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 values of x obtained from the circumvent above, the average result of x is 12.12. We discount 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 mo les 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%.DiscussionSeveral 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 heating, there was 0.2521 g of contents (KAl(SO4)2) left in the crucible. That content 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 concernin g 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 calculated result and the literature result reflected to efficiency and accuracy of the laboratory.Through a serial 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 s tepsThere are 0.019 moles in 0.5 g of Aluminum. Similarly, there are 0.036 moles in 2.01 g of potassium hydroxide. We apply 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 number of moles and alums molar mass malum= n M= 0.019 476.1 6= 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%.About 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 meaningful amount of alum was lost during filtration, suction and crystallization, because of the fact that the filter paper was not wet lavish and the crucibles were not dry enough due to short maximum heating time.References1. Franklin and Marshall College Chemistry 111/112 testing ground Manual, Fall 2012/Spring 2013, p. 39-41.