thesizing Alum Adapted from: Gilltie at IU-Kokomo and Neidig at Lebanon Valley C
ID: 1029208 • Letter: T
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thesizing Alum Adapted from: Gilltie at IU-Kokomo and Neidig at Lebanon Valley College Purpose: In this lab, we will show the dramatic changes that a chemical reactions can cause to a We will take aluminum a soft shiny metal and chemically transform it into alum, a salt. also use the stoichiometry of the reaction to obtain a percent yield. material. We will Background Information: used Alum is used for a lot of different processes, including the papermaking industry where it is to fill the pores in the cellulose structure. This process is known as sizing, a amount of bleeding that occurs when ink is used on paper. Another use of alum is in the food processing where it is used to pickle food. nd reduces the Today, we will try to transform aluminum foil into alum so we can pickle some cucumbers. This process is multistep, but we can write each step as a chemical reaction. The first step is to dissolve the foil in potassium hydroxide, a strong base. The equation is shown belovw 2A(s) +20H + 6H20() 2ACOH)+3H2(9) After the solid is completely dissolved, we will filter and add sulfuric acid, this will cause the production of aluminum hydroxide, which will be a solid. Al OH) (aa)+H2SOsCaq) ACOH), (s)+ HSO, (aq)+H2oD) This reaction will yield a solid, but it is not the alum we are hoping to produce. Additional will dissolve the aluminum hydroxide to give free aluminum ions. acid Al(OH)3 (s) + H30+ (aq) AP+ (aq) + 6H20(l) After heating and cooling, the atoms will have enough time to form our final product alum. Alum salts in general can have many formulas, the one we are creating today is KAI(SO4).12H20(s). When we calculate our molar mass, we need to take those 12 waters into consideration. Our overall reaction written with only the involved species, called a net ionic equation, is: Al (a)+ Kt(aq)+ 250, (aa)+ 12H20(0)-KASO)2.12H20() out a We can use the stoichiometry we have been p from the amount of aluminum foil we started with, and percent yield of product based on how much product we were able to isolate.Explanation / Answer
1. Al(OH)3 + NaOH + 2H2SO4 + 12H2O -----> NaAl(SO4)2.12H2O + 4H2O
2. From the sequence of ionic reactions given for the synthesis of alum, it can be observed that at first, two equivalents of aluminum are consumed to produce tetrahydroxide aluminum anion which gives a 1:1 ratio of alum. So if 0.456g of Al foil was taken, it will contain 0.456/26.9815 = 16.9004mmoles which is truly twice the moles of alum produced giving the theoretical yield of potash alum to be 8.4502mmoles. With the molar mass of hydrated potash alum as 474.3884g/mol, the theoretical yield of alum is 8.4502x10-3*474.3884 = 4.0086g.
NOTE: The higher than maximal yield of alum might be due to improper drying of the alum prepared thus having the leftover water contributing to an increase in weight.
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