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Alum Synthesis Lab Report

965 Words4 Pages

In order to properly appreciate the importance of aluminum recycling in our world, an aluminum recycling experiment was done through alum synthesis. The process by which this was done recycled solid aluminum can pieces (Al) into the form of solid raw alum crystals (KAl(SO4)2) through a series of reactions. Application of this experiment comes from analyzing the calculated percent yield of alum crystals, understanding where error is involved, and determining how a perfect yield could be obtained. First, 0.9-1.2 grams of aluminum pieces were weighed out and placed in a 250 mL beaker, in which 50 mL of 1.4 M KOH solution was added and a fume hood was placed. The mixture was placed on a hot plate and stirred to speed up the reaction. Meanwhile, …show more content…

2KAl(OH)4 (aq) + H2SO4 (aq) → 2Al(OH)3 (s) + K2SO4 (aq) + 2H2O (l)
In this acid-base/precipitate reaction, the sulfuric acid added to the potassium aluminum hydroxide formed aluminum hydroxide, the white, jello-consistency precipitate that was observed. Potassium sulfate and water were also formed.
3. 2Al(OH)3 (s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 6H2O (l)
Immediately after the precipitate was formed, the third, acid-base/metathesis reaction started. The solid aluminum hydroxide mixed with the sulfuric acid reforms into an aqueous aluminum sulfate in the water that was also formed. In this reaction, the aluminum sulfate created in the second reaction is being used up with the excess sulfuric acid in the third. This process was initiated as soon as the precipitate formed, but only time allowed for the formation of aluminum sulfate. Thus, the solution was heated up to speed up the reaction. Once this step was completed, the solution appeared smooth and clear again, representing the aqueous aluminum …show more content…

Al2(SO4)3 (aq) + K2SO4 (aq) + 24 H2O (l) → 2KAl(SO4)2 ∙ 12H2O (s)
In this combination reaction, aluminum sulfate, potassium sulfate, and water were combined to form potassium aluminum sulfate, or the white, cornstarch-consistency alum crystals in water. It was this water that had to be filtered out of the alum crystals in order for them to dry and accurately record their mass. The theoretically yield for this experiment, based off of the mass of aluminum pieces originally weighed out, was 19.0144 g, while our actual yield was 12.7222 g. This is a 66.91% yield. Ideally, of course is a 100% yield. Factors that may have caused this in our own experiment include, but may not be limited to: 1) There was barely any hydrogen gas being produced in the first step when we took it off the heat, but enough to infer that the reaction may not have been completely over, possibly affecting the amount of potassium aluminum hydroxide produced 2) Having to filter the reaction mixture in the first step more than once, because we neglected to turn on the vacuum the first time, we may have lost some of the mixture in the process. 3) In the last step, when the solution was put in the ice bath, it was up to us when we thought enough crystals had been formed as to when to filter the crystals. Perhaps waiting longer would have produced more

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