Question

Given the equation:

Answer 1

Answer: a)  $Fe$ is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

$4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al$  

To calculate the moles :

   

$\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles$

$\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles$

According to stoichiometry :

a) 9 moles of $Fe$  require= 4 moles of $Al_2O_3$

Thus 0.15 moles of $Fe$ will require=$\frac{4}{9}\times 0.15=0.067moles$  of $Al$

Thus $Fe$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

b) As 9 moles of $Fe$ give = 8 moles of $Al$

Thus 0.15 moles of $Fe$ give =$\frac{8}{9}\times 0.15=0.133moles$  of $Al$

Mass of $Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g$

c) As 9 moles of $Fe$ give = 3 moles of $Fe_3O_4$

Thus 0.15 moles of $Fe$ give =$\frac{3}{9}\times 0.15=0.05moles$  of $Fe_3O_4$

Mass of $Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g$