Many grams of aluminum are required to produce 3.5 moles Al2O3 in the presence of excess O2?

The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below

write the equation for reaction

4 Al + 3O2 = 2 Al2O3

by use of mole ratio between Al to Al2O3 which is 4 : 2 the moles of Al

=3.5 x4/2 = 7 moles

mass of Al = moles / x molar mass

= 7 moles x27 g/mol = 189 grams