When 26.5 mol of aluminum oxide (Al_2O_3) are decomposed according to the reaction 2 Al_2O_3 → 4 Al + 3 O_2, how many moles of oxygen gas (O_2) are produced?

• A. 26.5 mol
• B. 17.7 mol
• C. 53.0 mol
• D. 39.8 mol

Answer: (D)

Verified step by step guidance

1. Write down the balanced chemical equation: \[2\ Al_2O_3 \rightarrow 4\ Al + 3\ O_2\]
2. Identify the mole ratio between aluminum oxide (\[Al_2O_3\]) and oxygen gas (\[O_2\]) from the balanced equation. For every 2 moles of \[Al_2O_3\] decomposed, 3 moles of \[O_2\] are produced.
3. Set up a proportion to find the moles of \[O_2\] produced from 26.5 moles of \[Al_2O_3\] using the mole ratio: \[\frac{3\ mol\ O_2}{2\ mol\ Al_2O_3} = \frac{x\ mol\ O_2}{26.5\ mol\ Al_2O_3}\]
4. Solve for \[x\], the moles of \[O_2\] produced, by cross-multiplying and dividing: \[x = 26.5 \times \frac{3}{2}\]
5. Calculate the value of \[x\] to find the number of moles of oxygen gas produced.