Answer: a) [tex]Fe[/tex] is the limiting reagent
b) 3.59 g
c) 11.6g
Explanation:
[tex]4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al[/tex]
To calculate the moles :
[tex]\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles[/tex]
[tex]\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles[/tex]
According to stoichiometry :
a) 9 moles of [tex]Fe[/tex] require= 4 moles of [tex]Al_2O_3[/tex]
Thus 0.15 moles of [tex]Fe[/tex] will require=[tex]\frac{4}{9}\times 0.15=0.067moles[/tex] of [tex]Al[/tex]
Thus [tex]Fe[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
b) As 9 moles of [tex]Fe[/tex] give = 8 moles of [tex]Al[/tex]
Thus 0.15 moles of [tex]Fe[/tex] give =[tex]\frac{8}{9}\times 0.15=0.133moles[/tex] of [tex]Al[/tex]
Mass of [tex]Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g[/tex]
c) As 9 moles of [tex]Fe[/tex] give = 3 moles of [tex]Fe_3O_4[/tex]
Thus 0.15 moles of [tex]Fe[/tex] give =[tex]\frac{3}{9}\times 0.15=0.05moles[/tex] of [tex]Fe_3O_4[/tex]
Mass of [tex]Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g[/tex]
