Answer : The mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
Explanation : Given,
Mass of [tex]Al[/tex] = 26.1 Kg = 26100 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]Al_2O_3[/tex] = 32 g/mole
First we have to calculate the moles of [tex]Al[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles[/tex]
Now we have to calculate the moles of [tex]Al_2O_3[/tex].
The balanced chemical reaction is,
[tex]2Al+Fe_2O_3\rightarrow Heat+Al_2O_3+2Fe[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react to give 1 mole of [tex]Al_2O_3[/tex]
So, 967.38 moles of [tex]Al[/tex] react to give [tex]\frac{967.38}{2}=483.69[/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=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg[/tex]
Therefore, the mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
In the reaction of 26.1 kg of Al with an excess of Fe₂O₃, whose equation is 2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l), will be produced 49.31 kilograms of Al₂O₃.
The reaction is:
2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l) (1)
To find the mass of Al₂O₃ produced, we need to find the number of moles of Al since it is the limiting reactant (Fe₂O₃ is in excess).
[tex] n_{Al} = \frac{m_{Al}}{A_{Al}} [/tex] (2)
Where:
[tex]m_{Al}[/tex]: is the mass of Al = 26.1 kg = 26100 g
[tex]A_{Al}[/tex]: is the atomic mass of Al = 26.982 g/mol
The number of moles of Al is (eq 2):
[tex]n_{Al} = \frac{m_{Al}}{A_{Al}} = \frac{26100 g}{26.982 g/mol} = 967.31 \:moles[/tex]
From equation (1) we have that 2 moles of Al react with 1 mol of Fe₂O₃ to form 1 mol of Al₂O₃(s), so the number of moles of Al₂O₃ produced is:
[tex]n_{Al_{2}O_{3}} = \frac{1 \:mol \:Al_{2}O_{3}}{2 \:moles \:Al}*967.31 \:moles \:Al = 483.66 \:moles[/tex]
Finally, the mass of Al₂O₃ in kilograms is:
[tex]m = n_{Al_{2}O_{3}}*MM = 483.66 \:moles*101.96 \:g/mol = 49313.5 g = 49.31 kg[/tex]
Therefore, will be produced 49.31 kilograms of Al₂O₃.
Read more about limiting reactants here:
I hope it helps you!
