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2 Al + Fe2O3 ----> Al2O3 + 2 Fe
We know that Al2O3 is the chemical formula for aluminum oxide.
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
Al2O3.
The reaction is with aluminium:Fe2O3 + 2 Al = 2 Fe + Al2O3
Al2O3 on thermal decomposition gives Al & O2. 1 mole Al2O3 gives 2 mole Al. 102 kg Al2O3 gives 54kg Al 1 kg Al2O3 gives 0.512 kg Al.
so you find the ratio of Aluminum to Oxygenin this case it is 2:3and then because the total mole is 2.16for aluminum:2.16/5*2 = 0.864for oxygen:2.16/5*3 = 1.296hope this helped :D
Al and S combine to form aluminum sulfide, Al2S3
Mn2O7+Al=
2 Al + Fe2O3 ----> Al2O3 + 2 Fe
Al2O3
Fe2O3 + 2Al --> Al2O3 + 2Fe
Al+3 O-2 ---------> these are the ions with their chargesAl2O3 -------------> this is when they are bonded together Al2O3
Aluminum oxide is Al2O3
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
We know that Al2O3 is the chemical formula for aluminum oxide.
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)