Question

the chemical equation below shows the formation of aluminum oxide (al₂o₃) from aluminum (al) and oxygen (o₂).
4al + 3o₂ → 2al₂o₃
the molar mass of o₂ is 32.0 g/mol. what mass, in grams, of o₂ must react to form 3.80 mol of al₂o₃?
○ 60.8 grams
○ 81.1 grams
○ 122 grams
○ 182 grams

Explanation:

Step1: Find mole ratio of O₂ to Al₂O₃

From the equation \(4\text{Al} + 3\text{O}_2
ightarrow 2\text{Al}_2\text{O}_3\), the mole ratio of \(\text{O}_2\) to \(\text{Al}_2\text{O}_3\) is \(\frac{3\ \text{mol}\ \text{O}_2}{2\ \text{mol}\ \text{Al}_2\text{O}_3}\).

Step2: Calculate moles of O₂

Given moles of \(\text{Al}_2\text{O}_3 = 3.80\ \text{mol}\).
Moles of \(\text{O}_2 = 3.80\ \text{mol}\ \text{Al}_2\text{O}_3 \times \frac{3\ \text{mol}\ \text{O}_2}{2\ \text{mol}\ \text{Al}_2\text{O}_3}\)
\(= 3.80 \times \frac{3}{2} = 5.70\ \text{mol}\ \text{O}_2\).

Step3: Calculate mass of O₂

Molar mass of \(\text{O}_2 = 32.0\ \text{g/mol}\).
Mass of \(\text{O}_2 = \text{moles} \times \text{molar mass}\)
\(= 5.70\ \text{mol} \times 32.0\ \text{g/mol}\)
\(= 182.4\ \text{g} \approx 182\ \text{g}\) (rounded appropriately).

Answer:

182 grams