Question

Al (s) + Fe2O3 (s) --> Al2O3 (s) + Fe (s) (needs balancing)

Answer 1

Answer: 134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of $Fe_2O_3$.

Solution:

Given :$Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+Fe(s)$

After balancing, $2Al(s)+Fe_2O_3 (s)\rightarrow Al_2O_3(s)+2Fe(s)$

As we are given $Fe_2O_3$ is in excess. Hence, Al will be considered as limiting reagent because it limits the formation of product.

$\text{Number of moles of}Al=\frac{\text{mass of the compound}}{\text{Molecular mass of the compound}}=\frac{65.2 g}{26.98 g/mol}=2.4166 moles$

According to reaction ,2 moles of $Al$ reacts with one mole of $Fe_2O_3$ to give 2 moles of Fe then 2.4166 moles of $Al$ will give :

$=\frac{2}{2}\times 2.4166\text{moles of}Fe= 2.4166 moles$

Mass of $Fe$ produced =$\text{Number of moles of compound}\times \text{Molecular mass of the compound}=2.4166 moles\times 55.85g/mol=134.967 g$

134.967 grams of Fe will be produced when 65.2 g of Al is reacted with an excess (unlimited) supply of $Fe_2O_3$.