Question

In a crucible, a mixture of 15.325 g of iron (III) oxide and 25.734 g of aluminum metal is heated and the following reduction of the oxide takes place:

Fe₂O₃ (s) + 2 Al (s) --> 2 Fe (l) + Al₂O₃ (l)

a) What is the limiting reactant?

b) Determine the maximum amount of aluminum oxide in grams that can be produced.

c) Calculate the mass of excess reactant remaining in the crucible.

Answer 1

Fe₂O₃ (s) + 2 Al (s) 2 Fe (l) + Al₂O₃ (l)

Molar mass of Fe₂O₃ =(2xAt.mass of Fe) + (3xAt.mass of O)

                                = (2x55.8) + (3x16)

                                = 159.6 g/mol

Molar mass of Al= 27 g/mol

Molar mass of Al₂O₃ = (2xAt.mass of Al) + (3xAt.mass of O)

                               = (2x27) + (3x16)

                               = 102g/mol

According to the balanced equation ,

1 mole of Fe₂O₃ reacts with 2 moles of Al

                                OR

1x159.6 g of Fe₂O₃ reacts with 2x27 g of Al

15.325 g of Fe₂O₃ reacts with M g of Al

M = (15.325x2x27) / 159.6

    = 5.185 g

So 25.734 - 5.185 = 20.549 g of Al metal left unreacted so it is the excess reactant.

Since all the mass of Fe₂O₃ completed it is the limiting reactant.

From the balanced equation,

1 moles of Fe₂O₃ produces 1 mole of Al₂O₃

                                  OR

1x159.6 g of Fe₂O₃ produces 1x102 g of Al₂O₃

15.325 g of Fe₂O₃ produces N g of Al₂O₃

N = ( 15.325x1x102) / (1x159.6)

   = 9.794 g of Al₂O₃

Therefore the maximum amount of Al₂O₃ produced is 9.794 g