The combustion of liquid octane, C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water

The combustion of liquid octane, C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water. How many moles of water are produced by the reaction of 1.80 moles of octane? ————–mol H2O

The Correct Answer and Explanation is:

Let’s go step by step carefully.

1. Write the balanced chemical equation

The combustion of octane (C₈H₁₈) can be written as: C8H18(l)+O2(g)→CO2(g)+H2O(l)\text{C}_8\text{H}_{18}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)C8​H18​(l)+O2​(g)→CO2​(g)+H2​O(l)

Now we balance it.

• Each octane molecule has 8 carbon atoms → we need 8 CO₂ molecules.
• Each octane molecule has 18 hydrogen atoms → we need 9 H₂O molecules (since each water molecule has 2 hydrogens: 18 ÷ 2 = 9).
• Now, balance oxygen:
  • 8 CO₂ gives 8×2 = 16 oxygen atoms.
  • 9 H₂O gives 9×1 = 9 oxygen atoms.
  • Total oxygen atoms needed = 16 + 9 = 25.
  • Therefore, we need 25/2 = 12.5 O₂ molecules.

To eliminate the fraction, multiply the entire equation by 2: 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(l)2\text{C}_8\text{H}_{18}(l) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(l)2C8​H18​(l)+25O2​(g)→16CO2​(g)+18H2​O(l)

2. Use the mole ratio

From the balanced equation:
2 moles of C₈H₁₈ produce 18 moles of H₂O.

Thus, the mole ratio is: 18 mol H2O2 mol C8H18=9 mol H2O per mol C8H18\frac{18\ \text{mol H}_2\text{O}}{2\ \text{mol C}_8\text{H}_{18}} = 9\ \text{mol H}_2\text{O per mol C}_8\text{H}_{18}2 mol C8​H18​18 mol H2​O​=9 mol H2​O per mol C8​H18​

3. Calculate moles of water produced

Given:
1.80 moles of C₈H₁₈.

Using the ratio: 1.80 mol C8H18×9 mol H2O1 mol C8H18=16.2 mol H2O1.80\ \text{mol C}_8\text{H}_{18} \times \frac{9\ \text{mol H}_2\text{O}}{1\ \text{mol C}_8\text{H}_{18}} = 16.2\ \text{mol H}_2\text{O}1.80 mol C8​H18​×1 mol C8​H18​9 mol H2​O​=16.2 mol H2​O

Answer: 16.2 mol H₂O

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4. Explanation (300 words like a textbook)

The combustion of hydrocarbons is a chemical reaction where the hydrocarbon reacts with oxygen to produce carbon dioxide and water. In this problem, the hydrocarbon is octane (C₈H₁₈), which is commonly found in gasoline.

To solve stoichiometry problems involving combustion, the first step is to write a balanced chemical equation. For octane, the unbalanced reaction is: C8H18+O2→CO2+H2O\text{C}_8\text{H}_{18} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}C8​H18​+O2​→CO2​+H2​O

Balancing carbon first, there are 8 carbon atoms in one molecule of octane, requiring 8 CO₂ molecules. Next, balancing hydrogen, 18 hydrogen atoms require 9 H₂O molecules. Finally, balancing oxygen, we find that 8 CO₂ molecules contribute 16 oxygen atoms and 9 H₂O molecules contribute 9 oxygen atoms, totaling 25 oxygen atoms. Since each O₂ molecule has 2 oxygen atoms, we require 12.5 O₂ molecules. To avoid fractions, we multiply the entire equation by 2: 2C8H18+25O2→16CO2+18H2O2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}2C8​H18​+25O2​→16CO2​+18H2​O

The balanced equation shows that 2 moles of octane produce 18 moles of water, establishing a mole ratio of 9:1 (9 moles of H₂O per 1 mole of C₈H₁₈). Given 1.80 moles of octane, we multiply by this ratio: 1.80×9=16.2 moles of water1.80 \times 9 = 16.2 \text{ moles of water}1.80×9=16.2 moles of water

Thus, the complete combustion of 1.80 moles of octane produces 16.2 moles of water.