Question

the reaction between aluminum and iron(iii) oxide can generate temperatures approaching 3000c and is used in welding metals. In one process, 268g of Al are reacted with 501g of Fe2O3. identify the limiting reagent and calculate the theoretical mass

Answer 1

Answer:

- Iron (III) oxide is the limiting reactant.

- $m_{Al_2O_3}=319.9gAl_2O_3$

- $m_{Fe}=350.4gFe$

Explanation:

Hello,

In this case, we consider the following reaction:

$2Al + Fe_2O_3 \rightarrow Al_2O_3 +2Fe$

Thus, for identifying the limiting reactant we should compute the available  moles of aluminium in 268 g:

$n_{Al}=268gAl*\frac{1molAl}{26.98gAl} =9.93molAl$

Next, we compute the moles of aluminium that are consumed by 501 grams of iron (III) oxide via their 2:1 molar ratio:

$n_{Al}^{consumed}=501gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_30}*\frac{2molAl}{1molFe_2O_3}=6.27molAl$

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:

$m_{Al_2O_3}=6.27molAl*\frac{1molAl_2O_3}{2molAl} *\frac{101.96gAl_2O_3}{1molAl_2O_3} =319.9gAl_2O_3$

And the theoretical mass of iron is:

$m_{Fe}=6.27molAl*\frac{2molFe}{2molAl} *\frac{55.845 gFe}{1molFe} =350.4gFe$

Best regards.