In the reaction 4Al + 3O_2 → 2Al_2O_3, how many moles of O_2 are consumed when 11.0 mol of Al are reacted?

• A. 16.5 mol
• B. 5.50 mol
• C. 8.25 mol
• D. 3.67 mol

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: \$4Al + 3O_2 \rightarrow 2Al_2O_3$.
2. Determine the mole ratio between aluminum (Al) and oxygen gas (\(O_2\)) from the balanced equation. For every 4 moles of Al, 3 moles of \(O_2\) are consumed.
3. Set up a proportion to find the moles of \(O_2\) consumed when 11.0 moles of Al react: \(\frac{3\ \text{mol}\ O_2}{4\ \text{mol}\ Al} = \frac{x\ \text{mol}\ O_2}{11.0\ \text{mol}\ Al}\).
4. Solve for \(x\) (moles of \(O_2\)) by cross-multiplying and dividing: \(x = \frac{3}{4} \times 11.0\).
5. Calculate the value of \(x\) to find the moles of \(O_2\) consumed (do not provide the final numeric answer here).