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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.