# Question 56818

Jul 19, 2017

Here's what I got.

#### Explanation:

Notice that according to the balanced chemical equation

$2 {\text{Al"_ ((s)) + "Fe"_ 2"O"_ (3(s)) -> "Al"_ 2"O"_ (3(s)) + 2"Fe}}_{\left(s\right)}$

you need $2$ moles of aluminium in order to react with $1$ mole of iron(III) oxide and produce $1$ mole of aluminium oxide, i.e. you have a $2 : 1$ mole ratio between the two reactants and a $2 : 1$ mole ratio between aluminium and aluminium oxide.

Now, use the molar masses of the two reactants to convert the masses to moles

124 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "4.596 moles Al"

601 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69 color(red)(cancel(color(black)("g")))) = "3.764 moles Fe"_2"O"_3

So now you must ask yourself--do I have enough moles of aluminium to ensure that all the moles of iron(III) oxide react?

Use the aforementioned mole ratio to find out!

3.764 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * "2 moles Al"/(1color(red)(cancel(color(black)("moles Fe"_2"O"_3)))) = "7.528 moles Al"

So, in order for all the moles of iron(III) oxide to react, you need to have $7.528$ moles of aluminium oxide.

Since you know that

overbrace("7.528 moles Al")^(color(blue)("what you need")) " " > " " overbrace("4.596 moles Al")^(color(blue)("what you have"))

you can say that aluminium will act as the limiting reagent, i.e. it will be completely consumed before all the moles of iron(III) oxide will get the chance to take part in the reaction.

The reaction will consume $4.596$ moles of aluminium and produce

4.596 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Al"_2"O"_3)/(2color(red)(cancel(color(black)("moles Al")))) = "2.298 moles Al"_2"O"_3

To convert this to grams, use the molar mass of aluminium oxide

$2.298 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"_2"O"_3))) * "101.96 g"/(1color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = color(darkgreen)(ul(color(black)("234 g}}}}$

The reaction will also consume

4.596 color(red)(cancel(color(black)("moles Al"))) * ("1 mole Fe"_2"O"_3)/(2color(red)(cancel(color(black)("moles Al")))) = "2.298 moles Fe"_2"O"_3#

which means that you will be left with

${\text{3.794 moles Fe"_2"O"_3 - "2.298 moles Fe"_2"O"_3 = "1.496 moles Fe"_2"O}}_{3}$

To convert this to grams, use the molar mass of iron(III) oxide

$1.496 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Fe"_2"O"_3))) * "159.69 g"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = color(darkgreen)(ul(color(black)("239 g}}}}$

The answers are rounded to three sig figs.