Answer:
- Iron (III) oxide is the limiting reactant.
- [tex]m_{Al_2O_3}=319.9gAl_2O_3[/tex]
- [tex]m_{Fe}=350.4gFe[/tex]
Explanation:
Hello,
In this case, we consider the following reaction:
[tex]2Al + Fe_2O_3 \rightarrow Al_2O_3 +2Fe[/tex]
Thus, for identifying the limiting reactant we should compute the available moles of aluminium in 268 g:
[tex]n_{Al}=268gAl*\frac{1molAl}{26.98gAl} =9.93molAl[/tex]
Next, we compute the moles of aluminium that are consumed by 501 grams of iron (III) oxide via their 2:1 molar ratio:
[tex]n_{Al}^{consumed}=501gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_30}*\frac{2molAl}{1molFe_2O_3}=6.27molAl[/tex]
Thus, we notice there are less consumed moles of aluminium than available, for that reason, it is in excess; therefore, the iron (III) oxide is the limiting reactant.
Moreover, the theoretical mass of aluminium oxide is:
[tex]m_{Al_2O_3}=6.27molAl*\frac{1molAl_2O_3}{2molAl} *\frac{101.96gAl_2O_3}{1molAl_2O_3} =319.9gAl_2O_3[/tex]
And the theoretical mass of iron is:
[tex]m_{Fe}=6.27molAl*\frac{2molFe}{2molAl} *\frac{55.845 gFe}{1molFe} =350.4gFe[/tex]
Best regards.
