Consider the reaction: ___ Al + ___ O2 --> ___ Al2O3. When 81 g of Al are reacted with 64 g of O2, is the excess reactant and of this reactant remain after the reaction is complete.

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)

a chemical reaction happens

explanation:

b

explanation:

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