Answer:
mass Al needed = 189 grams
Explanation:
4Al + 3O₂ => 2Al₂O₃
From equation 4 moles Al gives 2 moles 2Al₂O₃. So, how many moles Al gives 3.5 moles 2Al₂O₃?
Using ratio and proportion ...
4 moles Al / x = 2 moles Al₂O₃ / 3.5 moles 2Al₂O₃
solving for 'x' => x = 4 moles Al x 3.5 moles 2Al₂O₃ / 2 moles 2Al₂O₃
x = 188.86 grams Al ≅ 189 grams Al
