Respuesta :
Answer : The mass of [tex]Al_2O_3[/tex] formed will be, 468.18 grams.
Solution : Given,
Mass of Al = 124 g
Mass of [tex]Fe_2O_3[/tex] = 601 g
Molar mass of Al = 27 g/mole
Molar mass of [tex]Fe_2O_3[/tex] = 160 g/mole
Molar mass of [tex]Al_2O_3[/tex] = 102 g/mole
First we have to calculate the moles of Al and [tex]O_2[/tex].
[tex]\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{124g}{27g/mole}=4.59moles[/tex]
[tex]\text{ Moles of }Fe_2O_3=\frac{\text{ Mass of }Fe_2O_3}{\text{ Molar mass of }Fe_2O_3}=\frac{601g}{160g/mole}=3.76moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]Al[/tex] react with 1 mole of [tex]Fe_2O_3[/tex]
So, 4.59 moles of [tex]O_2[/tex] react with [tex]\frac{4.59}{2}=2.295[/tex] moles of [tex]Fe_2O_3[/tex]
From this we conclude that, [tex]Fe_2O_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Al[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Al_2O_3[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]Al[/tex] react to give 2 mole of [tex]Al_2O_3[/tex]
So, 4.59 moles of [tex]O_2[/tex] react to give [tex]\frac{2}{2}\times 4.59=4.59[/tex] moles of [tex]Al_2O_3[/tex]
Now we have to calculate the mass of [tex]Al_2O_3[/tex]
[tex]\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3[/tex]
[tex]\text{ Mass of }Al_2O_3=(4.59moles)\times (102g/mole)=468.18g[/tex]
Therefore, the mass of [tex]Al_2O_3[/tex] formed will be, 468.18 grams.
