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Recycling Aluminum

Recycling Aluminum - Recycling Aluminum Results and...

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Recycling Aluminum 10/17/08
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Results and Discussion Purpose The purpose of the experiment was to scale down the given procedure of recycling aluminum in order to produce a theoretical yield of 23 grams of Alum. This recycling follows the reaction of: 2Al (s) +2KOH (aq) +H 2 SO 4(aq) +22H 2 O (l) 2KAl(SO 4 ) 2 12H 2 O+3H 2(g). In order to calculate the required amount of aluminum, we work backwards: 23.0g alum x 1 mol alum / 474.4g alum x 26.98g Al / 1 mol Al = 1.308g of Al Therefore, 1.308g of Al is needed to produce a theoretical yield of 23g of Alum. Calculations The mass of the aluminum sample actually used was 1.300g. According to this quantity, the theoretical yield of our reaction becomes 1.300g Al x 1 mol Al / 26.98g of Al x 474.4g alum / 1 mol alum = 22.86g alum Since the molar ratio of given aluminum to theoretical aluminum is calculated as: .0485 mol of Al / .445 mol of Al = .109 We can now find how much of each reagent is needed: KOH 0.420L x 2mol/1L = 0.84mol
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0.84mol (.109) = 0.0916mol 0.0916mol x 1L / 1.4mol = 65.4mL KOH H 2 SO 4 0.216L x 10mol/1L = 2.16mol 2.16mol (.109)= 0.235mol .235mol x 1L / 9mol = 26.1mL H 2 SO 4 For H 2 0 and the ethanol/water mixture we did not use the molar ratio on the moles of each.
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