Answer: 134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of [tex]Fe_2O_3[/tex].
Solution:
Given :[tex]Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+Fe(s)[/tex]
After balancing, [tex]2Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+2Fe(s)[/tex]
As we are given [tex]Fe_2O_3[/tex] is in excess. Hence, Al will be considered as limiting reagent because it limits the formation of product.
[tex]\text{Number of moles of}Al=\frac{\text{mass of the compound}}{\text{Molecular mass of the compound}}=\frac{65.2 g}{26.98 g/mol}=2.4166 moles[/tex]
According to reaction ,2 moles of [tex]Al[/tex] reacts with one mole of [tex]Fe_2O_3[/tex] to give 2 moles of Fe then 2.4166 moles of [tex]Al[/tex] will give :
[tex]=\frac{2}{2}\times 2.4166\text{moles of}Fe= 2.4166 moles [/tex]
Mass of [tex]Fe[/tex] produced =[tex]\text{Number of moles of compound}\times \text{Molecular mass of the compound}=2.4166 moles\times 55.85g/mol=134.967 g[/tex]
134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of [tex]Fe_2O_3[/tex].
