Question

Using the reaction, 4Al+3O2→2Al2O3, how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol O2 and 58.5g Al and 98.0g O2?

Answer 1

Answer:

Explanation:

The given reaction is:

4Al+3O2→2Al2O3

Illustration:

From the reaction

4 moles of Al require 3 mole of O₂

Hence, 1 mole of Al require (3/4) = 0.75 moles of O₂

Now if for each mole of Al, we require at least 0.75 moles of O₂ to react. If we have oxygen less than 0.75 moles, all oxygen will be consumed before aluminium and hence it will be limiting reactant.

Coming to question:

0.25 moles of Al will require 0.25 x 0.75 moles of oxygen = 0.1875 moles of oxygen.

We have 0.40 moles of O₂ which is more than sufficient and hence Al will be limiting reactant.

Mass of Al = 58.5 g

Number of moles of Al = 58.5/27 = 2.166 moles

Mass of O₂ = 98 g

Number of moles of O₂ = 98/32 = 3.0625 moles

Again, 2.166 moles of Al will require 0.75 x 2.166 moles of oxygen = 1.6245 moles of oxygen.

Available oxygen = 3.0625 moles which is more than sufficient and hence Aluminium will be limiting reactant.