Answer: 170 grams of [tex]Al_2O_3[/tex]
Explanation:
The balanced chemical reaction is:
[tex]4Al+3O_2(g)\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 3.33 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 3.33=2.49moles[/tex] of [tex]O_2[/tex]
Thus [tex]Al[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
As 4 moles of [tex]Al[/tex] give = 2 moles of [tex]Al_2O_3[/tex]
Thus 3.33 moles of [tex]Al[/tex] give =[tex]\frac{2}{4}\times 3.33=1.67moles[/tex] of [tex]Al_2O_3[/tex]
Mass of [tex]Al_2O_3=moles\times {\text {Molar mass}}=1.67moles\times 101.96g/mol=170g[/tex]
Thus 170 g of [tex]Al_2O_3[/tex] will be produced.
