Calculate how many grams of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al2O3). 4Al + 3O2 → 2Al2O3

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0.347

Explanation:

Al is the limiting reagent.

Explanation:

The stoichiometry of the reaction is such that 4 moles of Al are required for every 3 moles of diatomic oxygen. This means that if the ratio of Al to diatomic oxygen is greater than 4/3, then the oxygen is the limiting reagent. If the ratio of Al to diatomic oxygen is less than 4/3, then Al is the limiting reagent.

The problem state that the ratio of Al to diatomic oxygen is

0.32

0.26

=

16

13

=

48

39

<

52

39

=

4

3

The ration of Al to diatomic oxygen is less than 4/3, so Al is the limiting reagent.

We can also see this if we ask the question "How much Aluminum" is required to completely react 0.26 moles of diatomic oxygen??

0.26

moles

O

2

×

4

moles

A

l

3

moles

O

2

≈

0.347

moles

A

l

So it would take 0.347 moles of Al to completely react with all of the oxygen, however there is only 0.32 moles of Aluminum present, so there is not enough Al to react with all of the oxygen, and so we say that the Al is the limiting reagent.

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