The Aluminum in a 0.764 g. sample of an unknown material wasprecipitated as alum

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 %

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