What is the maximum number of moles of Al2O3 that can be produced by the reaction of. 4 mol of Al with. 4 mol of O2

Given:

Moles of Al = 0.4

Moles of O2 = 0.4

To determine:

Moles of Al2O3 produced

Explanation:

4Al + 3O2 → 2Al2O3

Based on the reaction stoichiometry:

4 moles of Al produces 2 moles of Al2O3

Therefore, 0.4 moles of Al will produce:

0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3

Similarly;

3 moles O2 produces 2 moles Al2O3

0.4 moles of O2 will yield: 0.4 * 2/3 = 0.267 moles

Thus Al will be the limiting reactant.

Ans: Maximum moles of Al2O3 = 0.2 moles

There can 0.2 moles Al2O3 be produced

Explanation:

Step 1: Data given

Moles of Al = 0.4 moles

Moles of O2 = 0.4 moles

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate the limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant. It will completely be consumed. (0.4 moles).

O2 is in excess. Therer ill react 3/4*0.4 = 0.3 moles O2

There will remain 0.4 - 0.3 = 0.1 mol

Step 4: Calculate moles of Al2O3

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 0.4 moles Al we'll have 0.4/2 = 0.2 moles Al2O3

There can 0.2 moles Al2O3 be produced