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| Chem Lab: Synthesis of an Alum KAl(SO4)2 x 12H2O |
Sliyah402 |
02/09/08 |
Hello I have two questions regarding this. During the experiment.5547g of aluminum foil were used (2.056x10-2 moles) 0.06 moles of KOH were used and 0.096 moles of H2S04 were used. Aluminum being the limiting reactant. the % yield of KAl(SO4)2 x 12H2O was determined to be 131.64% I know this can't be, although we did have A LOT of product form (actual yield was 8.083g). However the theo. yield may have been calculated wrong(the moles of alum were multiplied by the molecular weight of the alum). My next question is: what is the PHYSICAL property of the product (KAl(SO4)2 x 12H2O) that does not allow for the recovery of 100% of the product? any help would be appreciated. I understant the lab but I can't think of that physical property and I have No clue how I received a % yield of more than 100% |
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