Question

Given the equation:

4Al2O3 + 9Fe --> 3Fe3O4 + 8Al

If 27.5 g of Al2O3 reacted with 8.4 g of Fe, how many of Fe 3O4 are formed?

a) Calculate the limiting reactant


b) Calculate the number of grams of Al produced.


c) Calculate the number of grams of Fe3O4 produced.


d) Calculate the percent yield if 10g of Fe O4 were obtained?

Answer 1

Answer: a)
`Fe` is the limiting reagent

b) 3.59 g

c) 11.6g

Step-by-step explanation:

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

To calculate the moles :

`\text{Moles of} Al_2O_3=(27.5g)/(102g/mol)=0.27moles`

`\text{Moles of} Fe=(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=
`(4)/(9)* 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 =
`(8)/(9)* 0.15=0.133moles` of
`Al`

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

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

Thus 0.15 moles of
`Fe` give =
`(3)/(9)* 0.15=0.05moles` of
`Fe_3O_4`

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