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A 5.276 g sample of Al-Mg alloy is digested in nitric acid (to Al*3 and Mg"). The solution is diluted to 500.00 mL and three 100.00 mL aliquots are treated with ammonia, the resulting Al(OH)3 precipitate is dried and then ignited to alumina, Al2O3 (MM = 101.961 g/mol). If the average mass of alumina obtained was 1.444 g, what was the percentage of aluminum (MM = 26.9815 g/mol) in the original sample?

A 5.276 g sample of Al-Mg alloy is digested in nitric acid (to Al*3 and Mg"). The solution is diluted to 500.00 mL and three 100.00 mL aliquots are treated with ammonia, the resulting Al(OH)3 precipitate is dried and then ignited to alumina, Al2O3 (MM = 101.961 g/mol). If the average mass of alumina obtained was 1.444 g, what was the percentage of aluminum (MM = 26.9815 g/mol) in the original sample?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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How do I calculate % aluminum in this sample?
A 5.276 g sample of Al-Mg alloy is digested in nitric acid (to Al*3 and Mg*2). The solution is diluted to 500.00 mL and three 100.00 mL aliquots are treated with ammonia, the resulting Al(OH)3 precipitate is dried and then ignited to alumina, Al2O3 (MM = 101.961 g/mol). If the average mass of alumina obtained was 1.444 g, what was the percentage of aluminum (MM = 26.9815 g/mol) in the original sample?
Transcribed Image Text:A 5.276 g sample of Al-Mg alloy is digested in nitric acid (to Al*3 and Mg*2). The solution is diluted to 500.00 mL and three 100.00 mL aliquots are treated with ammonia, the resulting Al(OH)3 precipitate is dried and then ignited to alumina, Al2O3 (MM = 101.961 g/mol). If the average mass of alumina obtained was 1.444 g, what was the percentage of aluminum (MM = 26.9815 g/mol) in the original sample?
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