The Aluminum in a 0.764 g. sample of an unknown material wasprecipitated as alum
ID: 690028 • Letter: T
Question
The Aluminum in a 0.764 g. sample of an unknown material wasprecipitated as aluminum hydroxide,Al(OH)3, which was then converted toAl2O3 by heating strongly. If 0.126 g ofAl2O3 is obtained from the 0.764 g sample,what is the mass percent of aluminum in the sample? The Aluminum in a 0.764 g. sample of an unknown material wasprecipitated as aluminum hydroxide,
Al(OH)3, which was then converted toAl2O3 by heating strongly. If 0.126 g ofAl2O3 is obtained from the 0.764 g sample,what is the mass percent of aluminum in the sample?
Explanation / Answer
One mole of Aluminum gives one mole of aluminumhydroxide. When two moles of aluminum hydroxide areheated, 1 mole of aluminum oxide is obtained. . moles of Aluminum oxide obtained = mass/ molar mass = 0.126 g / 101.96 g/mol =0.00123 moles . so moles of Aluminum hydroxide = 0.0024 moles . This is equal to the moles of Aluminum present. Mass of Aluminum present = moles * atomic mass =0.0024 moles * 26.98 g/mol =0.067 g . Mass percent of Aluminum = ( mass of Aluminum / mass ofsample) * 100 =(0.067 g/ 0.764 g) * 100 =8.78 %Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.