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 | College Teacher | (Level 2) Distinguished Educator
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 | College Teacher | (Level 3) Assistant Educator
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 | College Teacher | (Level 1) Valedictorian
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
Now,
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
So
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.