8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol
= 2136 grams
It means that there are two molecules of Aluminum Oxide, Al2O3.
This is a synthesis reaction, and when it's balanced, it looks like this: 4Al + 3O2 --> 2Al2O3
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)
We know that Al2O3 is the chemical formula for aluminum oxide.
Al2O3 is aluminum oxide.
The ionic compound Al2O2 forms when aluminum reacts with oxygen.
2Al + 3O2 --> 2Al2O3 0.78 moles O2 (2 moles Al2O3/3 moles O2) 0.52 moles Al2O3 produced ===========================( assuming oxygen limits )
It means that there are two molecules of Aluminum Oxide, Al2O3.
This is a synthesis reaction, and when it's balanced, it looks like this: 4Al + 3O2 --> 2Al2O3
Aluminium + Oxygen ----> Aluminium oxideBalanced equation:4 Al + 3 O2 ----> 2 Al2O3
Fe2O3 + 2Al --> Al2O3 +2 Fe The reaction is commonly known as thermite for the enormous amount of heat produced. The iron produced by the reaction is molten.
When aluminum burns, the reaction is highly exothermic.
The balanced chemical equation for the reaction between Al and O2 is: 4Al + 3O2 -> 2Al2O3 Therefore, the coefficient needed for Al to balance the equation is 4.
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)
Burning of aluminum in oxygen (air): 4Al + 3O2 → 2Al2O3
4Fe + 3O2 -> 2Fe2O3 is the balanced equation.It'll need enough oxygen (> (0.46*3)/4 mole O2)to give (0.46*2)/4 mole Fe2O3, so 0.23 mole Fe2O3is produced.
Balanced equation: 4Al(s) + 3O2(g) ==> 2Al2O3(s)2.40 mol Al ==> 1.2 moles Al2O3 (mole ratio of Al2O3 : Al is 2:4 or 1:2)2.10 mol O2 ==> 1.4 moles Al2O3 (mole ratio of Al2O3:O2 is 2:3)Therefore, Al is the limiting reactant.Theoretical yield will thus be 1.2 moles Al2O3. If you need mass, it is 1.2 moles x 102 g/mol = 122 g