PRE-LAB CALCULATIONS Part II 1. A sample of alum is to be analyzed for aluminum

Question

PRE-LAB CALCULATIONS Part II 1. A sample of alum is to be analyzed for aluminum ion by precipitation of aluminum hydroxide following the procedure you will use. Calculate the percent aluminum in the alum for the following data: A 1.011-gram sample of alum was dissolved and analyzed. An excess of ammonia solution was added and the precipitate filtered appropriately into a piece of filter paper. The filter paper was heated in a 22.018 g crucible. The filter paper burned away to leave only aluminum oxide in the crucible. The aluminum oxide and crucible weighed. 22.127 g. Calculate the percent aluminum in the alum analyzed. Percent aluminum in product mnound notassium chromium sulfate

Explanation / Answer

Balanced reaction for alum with ammonia solution is

2 KAl(SO4)2*12H2O + 6 NH3 = 2 Al(OH)3 + 3 (NH4)2SO4 + 18 H2O + K2(SO4)

2 moles 2 moles

Molar mass of alum = 474.39 g/mol

Given Mass of alum = 1.011 g

No. of moles of alum = 1.011 g / 474.39 g/mol = 2.13 * 10-3 moles

2 moles of alum will give 2 moles of aluminium hydroxide

so 2.13 * 10-3 moles of alum will give 2.13 * 10-3 moles of aluminium hydroxide

Aluminium hydroxide will decompose according to the following reaction

2 Al(OH)3 = Al2O3 + 3 H2O

2moles of Al(OH)3 will give 1 mole of oxide

so 2.13 * 10-3 moles of  Al(OH)3 will give 2.13 * 10-3 /2 = 1.06 * 10-3 moles of oxide

Molar mass of aluminium oxide = 101.96 g/mol

Mass of Al2O3 = 1.065 * 10-3 moles * 101.96 g/mol = 0.109 g Theoretical yield

Given Mass of crucible = 22.127 g

Mass of crucible + aluminium = 22.018 g

Mass of aluminium = 22.127 g - 22.018 g = 0.109 g

Each mole of Al2O3 will contain 2 moles of Al

so 1.06 * 10-3 moles of Al2O3 will have 2.13 * 10-3 moles of Al

Molar mass of Al = 27 g/mol

Mass of Al = 2.13 * 10-3 moles * 27 g/mol = 0.05751 g

Mass percent of aluminium in product = 0.05751 g / 0.109 g * 100 % = 52.76 %

mass of Al = 2.13 * 10-3 moles *

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