If #"2.112 g Fe"_2"O"_3"# reacts with #"0.687 g Al"#, what is the maximum amount of #"Fe"# that can be produced?

1 Answer
Meave60
Nov 3, 2015

Aluminum is the limiting reactant. The maximum amount of pure #"Fe"# that can be produced by this reaction is #"1.422 g"#.

Explanation:

#"Fe"_2"O"_3("s")" + 2Al(s)"##rarr##"Al"_2"O"_3("s")" + 2Fe(s)"#

We need to find the limiting reactant (also called limiting reagent).

Iron(III) Oxide

  1. First divide the given mass of #"Fe"_2"O"_3"# by its molar mass of #"159.687 g/mol"#.

  2. Next multiply times the mole ratio of #"Fe"# and #"Fe"_2"O"_3"# from the balanced equation.

  3. Then multiply times the molar mass of #"Fe"#, which is #"55.845 g/mol"#.

  4. This will give you the amount of pure #"Fe"# that can be produced from #"2.112 g Fe"_2"O"_3"#.

#2.112cancel"g Fe"_2"O"_3xx(1cancel"mol Fe"_2"O"_3)/(159.687cancel"g Fe"_2"O"_3)xx(2cancel"mol Fe")/(1cancel"mol Fe"_2"O"_3)xx(55.845"g Fe")/(1cancel"mol Fe")="1.4772 g Fe"#

Aluminum

  1. First divide the given mass of #"Al"# by its molar mass of #"26.9815 g/mol"#.

  2. Next multiply times the mole ratio of #"Fe"# and #"Al"# from the balanced equation.

  3. Then multiply times the molar mass of #"Fe"#, which is #"55.845 g/mol"#.

  4. This will give you the amount of pure #"Fe"# that can be produced from #"0.687 g Al"#.

#0.687cancel"g Al"xx(1cancel"mol Al")/(26.9815cancel"g Al")xx(2cancel"mol Fe")/(2cancel"mol Al")xx(55.845"g Fe")/(1cancel"mol Fe")="1.422 g Fe"#

Aluminum is the limiting reactant. The maximum amount of pure #"Fe"# that can be produced by this reaction is #"1.422 g"#.

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