Respuesta :
Answer:
34.23 grams is the maximum amount of aluminum oxide that can be formed.
Iron (III) oxide is a limiting reagent i.e [tex]Fe_2O_3[/tex].
4.6764 grams is the amount of the aluminum which remains after the reaction is complete
Explanation:
iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)
[tex]Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)[/tex]
Moles of iron(III) oxide : [tex]\frac{53.7 g}{160 g/mol}=0.3356 mol[/tex]
Moles of aluminium : [tex]\frac{22.8 g}{27 g/mol}=0.8444 mol[/tex]
According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.
Then 0.3356 moles of iron(III) oxide will react with:
[tex]\frac{2}{1}\times 0.3356 mol=0.6712 mol[/tex] of aluminum.
As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e [tex]Fe_2O_3[/tex] and aluminum in the an excessive reagent.
Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.
According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.
Then 0.3356 moles will give:
[tex]\frac{1}{1}\times 0.3356 mol=0.3356 mol[/tex] of aluminum oxide
Mass of 0.3356 moles of aluminum oxide:
0.3356 mol × 102 g/mol = 34.23 g
34.23 grams is the maximum amount of aluminum oxide that can be formed.
Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol
Mass of 0.1732 moles of aluminum :
0.1732 mol × 27 g/mol = 4.6764 g
4.6764 grams is the amount of the aluminum which remains after the reaction is complete
