4A1 + 302 – 2Al2O3 How many grams of aluminum(Al) are required to produce 3.5 moles Al2O3 in the
presence of excess O2?
94.5 g
189 g
378 g
108 g

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

the answer is b. dilution