How many grams of al2o3 can be made by reacting 4.6 grams of al with excess of o2? (always balance the equation) al + o2 > al2o3

The answer to your question is 8.67 grams of Al₂O₃

Explanation:

Data

mass of Al₂O₃ = ?

mass of Al = 4.6 g

mass of O₂ = excess

Balanced Reaction

                   4Al   +    3O₂     ⇒    2Al₂O₃

            Reactants           Elements           Products

                    4                     Al                          4

                    6                      O                         6

Process

1.- Use proportions to calculate the moles of Al

                         27 g of Al  1 mol

                         4.6 g of Al  x

                           x = 0.17 mol of Al

2.- use proportions to calculate the moles of Al₂O₃

                   4 moles of Al  2 moles of Al₂O₃

                   0.17 moles of Al  x

                        x = 0.085 moles of Al₂O₃

3.- Use proportions to calculate the grams of Al₂O₃

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

                     102 g of Al₂O₃  1 mol

                       x                     0.085 moles

                       x = 8.67 g of Al₂O₃

c) should be the answer.