What mass of aluminum oxide can be recovered from the complete reaction of 31.5 g of Al with 31.5 g of oxygen gas? 4 Al (s) + 3 O2(g) → 2 Al2O3(s)

The answer to your question is  59.5 g

Explanation:

Chemical reaction

                             4Al   +   3O₂    ⇒     2Al₂O₃

Data

mass of Al₂O₃ = ?

mass of Al = 31.5 g

mass of O₂ = 31.5 g

Atomic mass Al = 4 x 27 = 108 g

Atomic mass O₂ = 6 x 16 = 96 g

Process

1.- Calculate the limiting reactant

Theoretical proportion   Al/O₂  = 108/96 = 1.125

Experimental proportion Al/O₂ = 31.5/31.5 = 1

The limiting reactant is Al because the proportion diminished

2.- Calculate the mass of Al₂O₃

molecular mass Al₂O₃ = (27 x 2) + (16 x 3) = 2(102 g) = 204 g

                   108 g  of Al 204 g of Al₂O₃

                   31.5 g of Al    x

                    x = (31.5 x 204)/108

                    x = 59.5 g of Al₂O₃  

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.