Chem Lab 4

Chem Lab 4 - Recycling Aluminum by Alex Wittenberg Lab...

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Recycling Aluminum by Alex Wittenberg Lab Instructor: Saba Khan October 2, 2009
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Results and Discussion : The purpose of this lab was to synthesize alum from aluminum by means of gravity and vacuum filtrations. The melting point of the solid alum was also determined. It was necessary to scale down the given procedure in order to prepare a theoretical yield of 23 g of alum. To find the amount of aluminum necessary to yield 23g: 23 g alum x 1 mole alum 690.5848 g alum x 2 mole Al 2 mole alum x 26.98 g Al 1 mole Al =1.31 g Al To find the factor that the Al must be reduced: 1.31 g Al x 1 mole Al 26.98 Al = 0.0485 mole Al ® 0.445 mole Al 0.0485 mole Al = 9.18 To find volume of KOH needed: 0.840 mole *0.109 = 0.0916 mole * 1 L 1.4 mole x 1000 mL 1 L = 65.4 mL To find volume of H 2 SO 4 needed: 2.16 mole *0.109 = 0.235 mole * 1 L 9.0 mole x 1000 mL 1 L = 26.16 mL To find volume of aliquot of sulfuric acid needed: 54 mL * 1 L 1000 mL x 10 mole 1 L x 0.109 x 1 L 9 mole x 1000 ml 1 L = 6.54 mL
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This note was uploaded on 10/29/2009 for the course CHEM 2070 taught by Professor Chirik,p during the Fall '05 term at Cornell.

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Chem Lab 4 - Recycling Aluminum by Alex Wittenberg Lab...

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