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?

Nov 3, 2015

Aluminum is the limiting reactant. The maximum amount of pure $\text{Fe}$ that can be produced by this reaction is $\text{1.422 g}$.

Explanation:

$\text{Fe"_2"O"_3("s")" + 2Al(s)}$$\rightarrow$$\text{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 $\text{Fe"_2"O"_3}$ by its molar mass of $\text{159.687 g/mol}$.

2. Next multiply times the mole ratio of $\text{Fe}$ and $\text{Fe"_2"O"_3}$ from the balanced equation.

3. Then multiply times the molar mass of $\text{Fe}$, which is $\text{55.845 g/mol}$.

4. This will give you the amount of pure $\text{Fe}$ that can be produced from $\text{2.112 g Fe"_2"O"_3}$.

$2.112 \cancel{\text{g Fe"_2"O"_3xx(1cancel"mol Fe"_2"O"_3)/(159.687cancel"g Fe"_2"O"_3)xx(2cancel"mol Fe")/(1cancel"mol Fe"_2"O"_3)xx(55.845"g Fe")/(1cancel"mol Fe")="1.4772 g Fe}}$

Aluminum

1. First divide the given mass of $\text{Al}$ by its molar mass of $\text{26.9815 g/mol}$.

2. Next multiply times the mole ratio of $\text{Fe}$ and $\text{Al}$ from the balanced equation.

3. Then multiply times the molar mass of $\text{Fe}$, which is $\text{55.845 g/mol}$.

4. This will give you the amount of pure $\text{Fe}$ that can be produced from $\text{0.687 g Al}$.

$0.687 \cancel{\text{g Al"xx(1cancel"mol Al")/(26.9815cancel"g Al")xx(2cancel"mol Fe")/(2cancel"mol Al")xx(55.845"g Fe")/(1cancel"mol Fe")="1.422 g Fe}}$

Aluminum is the limiting reactant. The maximum amount of pure $\text{Fe}$ that can be produced by this reaction is $\text{1.422 g}$.