To determine how many moles of water are produced when 1.0 mole of octane (C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex]) reacts with excess oxygen, let's follow these steps:
1. Write down the balanced chemical equation:
[tex]\[
2 \text{C}_8\text{H}_{18} + 25 \text{O}_2 \rightarrow 16 \text{CO}_2 + 18 \text{H}_2\text{O}
\][/tex]
2. Identify the stoichiometric ratios:
From the balanced chemical equation, we see the stoichiometric coefficients are as follows:
- 2 moles of octane (C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex])
- 25 moles of oxygen (O[tex]\(_2\)[/tex])
- 16 moles of carbon dioxide (CO[tex]\(_2\)[/tex])
- 18 moles of water (H[tex]\(_2\)[/tex]O)
3. Set up the ratio for octane and water:
According to the stoichiometric coefficients:
- 2 moles of C[tex]\(_8\)[/tex]H[tex]\(_{18}\)[/tex] produce 18 moles of H[tex]\(_2\)[/tex]O.
4. Find the ratio for 1 mole of octane:
Since 2 moles of octane produce 18 moles of water, we can calculate how many moles of water are produced by 1 mole of octane:
[tex]\[
\text{Moles of H}_2\text{O} = \left(\frac{18 \text{ moles H}_2\text{O}}{2 \text{ moles C}_8\text{H}_{18}}\right) \times 1 \text{ mole C}_8\text{H}_{18}
\][/tex]
5. Calculate the moles of water produced:
[tex]\[
\text{Moles of H}_2\text{O} = 9 \text{ moles H}_2\text{O}
\][/tex]
Therefore, when 1.0 mole of octane reacts with excess oxygen, 9.0 moles of water are produced.
