Aluminum reacts with oxygen in the following chemical reaction:  Al + O2 → Al2O3  If 3.17 g of Al and 2.55 g of O2 are available to react, which is the limiting reactant?

3 Answers

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The limiting reactant in a chemical equation is the reactant that is completely consumed when the chemical reaction takes place. As a result the reactant determines for how long the reaction takes place.

The reaction given is that between aluminum and oxygen to yield Al2O3. The equation of this chemical reaction is:

4Al + 3O2 --> 2Al2O3

3.17 g of Al and 2.55 g of O2 are available for the reaction to take place. The molar mass of Al is 27 and that of O2 is 32.

3.17 g of Al is equivalent to 0.11 moles and 2.55 g of O2 is equivalent to 0.079 moles

As 4 moles of Al react with 3 moles of oxygen in the reaction, there should be 3/4 times as much oxygen as aluminum. But the number of moles of the reactants available to react shows that a lesser amount of oxygen is available.

In this case, oxygen is limiting reagent and the reaction stops when all the oxygen is consumed.

sanjeetmanna's profile pic

sanjeetmanna | College Teacher | (Level 3) Assistant Educator

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First step balancing the given reaction.

4Al + 3O2 -----> 2Al2O3

Second step Find moles from the given mass.

3.17 gm of Al

Moles = mass / molar mass

Moles = 3.17 / 27

Moles = 0.11

2.55 gm of O2

Moles = mass / molar mass

Moles = 2.55 / 32

Moles = 0.07.

Moles of Al = 0.11,

Moles of O2 = 0.07

Third step Divide the Moles with the stoichiometric coefficient in the given chemical reaction.

For Al = 0.11/4 = 0.0275

For O2 = 0.0233

From the above two value we can see that O2 has less value that is 0.0233, so O2 is the limiting reactant for this reaction.

chaobas's profile pic

chaobas | College Teacher | (Level 1) Valedictorian

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Let's find the number of moles of Al and O2.

Mass of Al =3.17 g and molar mass of Al = 27 g/mol

Mass of O2 = 2.55 g and molar mass of O2 = 32 g/mol



Moles = mass/molar mass

Moles of Al = 3.17 g/ (27 g/mol) =0.117 mol Al

Moles of O2 = 2.55 g / (32 g/mol) = 0.080 mol O2



 we need to balanced the reaction.

4 Al + 3 O2 → 2 Al2O3  


4 moles of Al reacts with 3 moles of O2. That is the molar ratio is 4 Al: 3 O2

But we have 0.117 mol Al and 0.080 mol O2.

Let's calculate the nuimber of moles of Al requries by 0.080 mol O2

0.080 mol O2 * (4 mol Al/ 3 mol O2)

0.106 mol Al

But we have 0.117 mol Al. so there are plenty of Al.

Limiting reagent is the reagent which are completely utilized in the reaction and which determine the amount of product formed.

So in our case the reactant which is completely utilized would be O2.

So oxygen is the limiting reagent.