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?

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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.

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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.

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