In the reaction C_3H_8 + 5 O_2 → 3 CO_2 + 4 H_2O, how many moles of water (H_2O) will be produced if 1.2 mol of oxygen (O_2) reacts with excess propane (C_3H_8)?

• A. 2.4 mol
• B. 6.0 mol
• C. 0.96 mol
• D. 1.2 mol

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: \(\mathrm{C_3H_8 + 5\ O_2 \rightarrow 3\ CO_2 + 4\ H_2O}\).
2. Note the mole ratio between oxygen (\(\mathrm{O_2}\)) and water (\(\mathrm{H_2O}\)) from the balanced equation. For every 5 moles of \(\mathrm{O_2}\), 4 moles of \(\mathrm{H_2O}\) are produced.
3. Set up a proportion to find the moles of water produced from 1.2 moles of oxygen: \(\frac{4\ \mathrm{mol\ H_2O}}{5\ \mathrm{mol\ O_2}} = \frac{x\ \mathrm{mol\ H_2O}}{1.2\ \mathrm{mol\ O_2}}\).
4. Solve for \(x\) (moles of water) by cross-multiplying and dividing: \(x = \frac{4}{5} \times 1.2\).
5. Interpret the result as the number of moles of water produced when 1.2 moles of oxygen react with excess propane.