Aluminum cookware is reffered ti as "anodized aluminum". The anodizing process puts a layer of aluminum oxide on the aluminum, protecting it from corrosion. Consider the reaction of 10.00 grams of aluminum with 10.00 grams of oxygen to form aluminum oxide. What is the formula of aluminum oxide? Write a balance chemical equation for the reaction. Determine which reactant is the limiting reagent. How many grams of aluminum oxide will form? How many grams of excess reagent are left?

Question

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Answer 1

The reaction for the formation of aluminum oxide is:
4Al + 3O2 -->2 Al2O3
Moles of Al = 10.00g Al / 26.98 g/mol = 0.3706 mol
Moles of O2 = 10.00 g O2 / 31.999 g/mol = 0.3125 mol O2.
(mol. Al)/(mol.O2) = 0.3706 / 0.3125 = 1.186 (based on amounts used)
(mol. Al)/(mol.O2) = 4/3 = 1.333 (based on the chemical equation)
Since the amount of Al needed is LESS that the amount available, aluminum is the LIMITING REAGENT.
The calculation of the amount of Al2O3 MUST be based on the limiting reagent. Al. Now you set up:
(0.3706 mol Al)(2 mol Al2O3 / 4 mol Al) = 0.1853 moles Al2O3.
Last step: Convert moles to grams.

Answer 2

Sorry for a bad sentence in my answer:
"Since the amount of Al needed is LESS that the amount available, aluminum is the LIMITING REAGENT. ".
It should read:
Since the amount of Al needed is MORE that the amount available, aluminum is the LIMITING REAGENT.