If you have 108 grams of Al and 108 grams of O subscript 2 which reactant is the limiting reactant?

1. Write the balanced equation.

4Al + 3O2 ==> 2Al2O3.

2a. Convert 108 g Al to moles. moles = g/molar mass.
2b. Convert 108 g Oxygen to moles. same process.

3a. Using the coefficients in the balanced equation, convert moles Al in 2a to moles of the product.
3b. Using the same process, convert moles O2 in 2b to moles of the product.
3c. You will likely obtain two different answers. Of course, both CAN'T be correct. The correct answer is ALWAYS the smaller one. The smaller one in this case will be the oxygen and that is the limiting reagent.

4. The problem doesn't ask for it but you can determine the grams Al2O3 by using smaller value from 3c and convert it to grams Al2O3. grams = moles x molar mass.