The amount of [tex]Al_2O_3[/tex] that would be formed will be 44.59 grams
Stoichiometric calculation
From the equation of the reaction:
[tex]4Al + 3O_2 --- > 2Al_2O_3[/tex]
The mole ratio of Al to [tex]Al_2O_3[/tex] is 2:1
Mole of 23.6 g Al = 23.6/26.98 = 0.87 moles
Equivalent mole of [tex]Al_2O_3[/tex] = 0.87/2 = 0.44 moles
Mass of 0.44 mole [tex]Al_2O_3[/tex] = 0.44 x 101.96 = 44.59 grams
More on stoichiometric calculations can be found here: https://brainly.com/question/27287858
#SPJ1
