Answer:
Al is the excess reactant and 0.33 moles of this reactant remain after the reaction is complete
Explanation:
First of all, let's balance the equation:
4Al + 3O₂ → 2Al₂O₃
We need to know the moles of each reactant, to work with the stoichiometry.
Moles of Al: 81 g / 26.98 g/mol = 3 moles
Moles of O₂: 64 g / 32g/mol = 2 moles
If 4 moles of Al react to 3 moles of oxygen
3 moles of Al, may react with (3 . 3) / 4 = 2.25 moles of O₂
As we have 2 moles of oxygen and we need 2.25 moles, we do not have enough moles, so the O₂ is the limiting reactant. Then, the Al is the excess reactant.
3 moles of O₂ react to 4 moles of Al
Then, 2 moles of O₂ may react to (2 .4) / 3 = 2.67 moles of Al
We have 3 moles, and we need 2.67 moles, so there are 0.33 moles that still remain after the reaction goes complete (3 moles, we have - 2.67 moles we need)
