If aluminum (Al), with an atomic weight of 27, combines with oxygen (O), with an atomic weight 16, to form the compound aluminum oxide (Al2O3), how much oxygen would be required to react completely with 57g of aluminum ? Answer is in grams.

Answer 1

mass of mol of Al2O3 2*27+3*16 = 102 grams/mol
57 grams Al * 1 mol/27 grams = 2.11 mols
4Al + 3O2 --- > 2Al2O3
so we need 3 mols of O2 for every 4 moles of Al
(3 mols O2/4 mols Al ) * 2.11 mols Al = 1.58 mols O2
O2 is 32 grams/mol
so 1.58 mols * 32 grams / mol