Answer:
To solve this problem, we need to use stoichiometry to relate the amount of O2 used to the amount of Al2O3 produced, using the balanced chemical equation:
4 Al + 3 O2 → 2 Al2O3
First, we need to determine the limiting reactant between O2 and Al. To do this, we can calculate the amount of moles of each reactant:
moles O2 = mass ÷ molar mass = 64.0 g ÷ 32 g/mol = 2.00 mol
moles Al = 0 g ÷ 26.98 g/mol = 0 mol
Since there is no Al, it is the limiting reactant. Therefore, all of the O2 will react with 4 moles of Al to form 2 moles of Al2O3.
Now, we can calculate the amount of Al2O3 produced:
moles Al2O3 = 2 mol Al2O3 ÷ 4 mol Al × 2.00 mol O2 = 1.00 mol Al2O3
mass Al2O3 = moles Al2O3 × molar mass Al2O3 = 1.00 mol × 102 g/mol = 102 g
Therefore, 102 g of Al2O3 will form from 64.0 g O2.