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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?

User OSKM
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1 Answer

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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

User Jan Ariel San Jose
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8.1k points