Respuesta :

jufsanabriasa

Answer:

50.6g Al would be needed

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of aluminium react with 3 moles of oxygen to produce 2 moles of Al₂O₃

To solve this question we need to find the moles of 45g of O₂. Then using the chemical equation find the moles of Al and its mass:

Moles O₂ -Molar mass: 32g/mol-:

45g O₂ * (1mol / 32g) = 1.41 moles O₂

Moles Al:

1.41 moles O₂ * (4mol Al / 3mol O₂) = 1.875 moles Al

Mass Al -Molar mass: 26.98g/mol-:

1.875 moles Al * (26.98g / mol) =

50.6g Al would be needed

Oseni

The amount of aluminum that would be required to completely react with 45 grams of O2 would be 50.59 grams.

From the balanced equation of the reaction:

[tex]4 Al + 3 O_2 ---> 2 Al_2O_3[/tex]

The mole ratio of Al to O2 is 4:3.

Mole of 45 grams of O2 = mass/molar mass

                                     = 45/32

                                           = 1.4063 moles

Equivalent mole of Al: 4 x 1.4063/3

                                    = 1.875 moles

mass of Al = mole x molar mass

                 = 1.875 x 26.98

                  = 50.59 grams

More on stoichiometry calculations can be found here: https://brainly.com/question/22288091?referrer=searchResults