the grams of O2 required to react with 9.3 0 mole Al is calculated as followsl
that is 4Al + 3O2 = 2 Al2O3
from the reacting ratio of Al to O2 which is 4 :3 the moles of O2 = 9.30 x 3/4 = 6.975 moles
mass of O2 is therefore = moles x molar mass
= 6.975 moles x 32g/mol = 223.2 grams