Answer: 28.144 g of [tex]Al_2O_3[tex] will be produced.
Explanation: We are given that Al reacts with [tex]Fe_2O_3[/tex], then the balanced chemical reaction is:
[tex]2Al(s)+Fe_2O_3(aq.)\rightarrow Al_2O_3(aq.)+Fe(s)[/tex]
It is given that [tex]Fe_2O_3[/tex] is present in excess, so it is an excess reagent and Al is the limiting reagent.
Now, 2 moles of Al produces 1 mole of [tex]Al_2O_3[/tex], which means that (2×27g) = 54g of Al produces 102g of [tex]Al_2O_3[/tex].
Therefore, 14.9g of Al will produce = [tex]\frac{102g\times 14.9g}{54g}\text{ of }Al_2O_3[/tex]
= 28.144g of [tex]Al_2O_3[/tex]
