4 Al + 3 O2 -> 2 Al2O3 How much aluminum would be needed to completely react with 45 grams of Oz?

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