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ed chemical equation. However, the balanced chemical equation only works in units of moles. To convert grams of aluminum to moles of aluminum the atomic mass of aluminum must be used. grams of aluminum x 1 mole Al = 1 26.98 g Al To relate the moles of aluminum to KAl(SO4)2* 12H20, the mole to mole ratio must be found in the balanced chemical equation. In this case it is: 1 mole of Al3 = 1 mole KAl(SO4)2 12H20 As a conversion factor it can be written in one of two ways: 1 mole KAJ(SO4)2 12H2O 1 mole of Al*3 1 mole of Al3 1 mole KAI(SO4)2 or 12H20 Use dimensional analysis to cancel units of moles of aluminum to moles of KAI(SO4)2 12H2O. 12H20 1 mole Al x 1 mole KAl(SO42 1 mole of Alt3 12H20 = 1 mole of KAl(SO4)2 1 Finally, moles of KAl(SO4)2 12H20 can be converted to grams of KAl(SO4)2 by calculating the formula mass. 12H20 1 mole KAl(SO42 12H20 x Formula mass KA|(SO42 12H O= theoretical yield 1 mole KAl(SO4)2 12H20 1 Don't forget to include the mass of the water in the formula mass.
Data Table Measurements 1. Mass of aluminum l6.84 2. Mass of final product, KAl(SO4)2 12H20 Calculations (show all work on the next page) 3. Atomic mass of aluminum (units of grams/mole) 4. Moles of aluminum 5. Moles of KAl(SO4)2 12H20 6. Formula mass of KAl(SO4)2 12H20 (units of grams/mole from periodic table) 7. Theoretical mass of KAl(SO4)2* 12H20 (units of grams) 8. Percent yield Lab Problems
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