Certainly! Let's address this step-by-step:
1. Identify the balanced chemical equation for the combustion of propane:
[tex]\[
\text{C}_3\text{H}_8(g) + 5 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) + 4 \text{H}_2\text{O}(g)
\][/tex]
2. Determine the given information:
- Volume of propane consumed: [tex]\(2.50\)[/tex] liters
3. Identify the stoichiometric relationships from the balanced equation:
- 1 mole of [tex]\( \text{C}_3\text{H}_8 \)[/tex] reacts with 5 moles of [tex]\( \text{O}_2 \)[/tex]
- 1 mole of [tex]\( \text{C}_3\text{H}_8 \)[/tex] produces 3 moles of [tex]\( \text{CO}_2 \)[/tex]
- 1 mole of [tex]\( \text{C}_3\text{H}_8 \)[/tex] produces 4 moles of [tex]\( \text{H}_2\text{O} \)[/tex]
4. Convert the volume of propane to volumes of oxygen, carbon dioxide, and water vapor using stoichiometric coefficients:
- Oxygen Consumed:
According to the stoichiometry, 1 volume of [tex]\( \text{C}_3\text{H}_8 \)[/tex] requires 5 volumes of [tex]\( \text{O}_2 \)[/tex]. Thus,
[tex]\[
\text{Volume of } \text{O}_2 = 2.50 \text{ L of } \text{C}_3\text{H}_8 \times 5 = 12.5 \text{ L}
\][/tex]
- Carbon Dioxide Produced:
According to the stoichiometry, 1 volume of [tex]\( \text{C}_3\text{H}_8 \)[/tex] produces 3 volumes of [tex]\( \text{CO}_2 \)[/tex]. Thus,
[tex]\[
\text{Volume of } \text{CO}_2 = 2.50 \text{ L of } \text{C}_3\text{H}_8 \times 3 = 7.5 \text{ L}
\][/tex]
- Water Vapor Produced:
According to the stoichiometry, 1 volume of [tex]\( \text{C}_3\text{H}_8 \)[/tex] produces 4 volumes of [tex]\( \text{H}_2\text{O} \)[/tex]. Thus,
[tex]\[
\text{Volume of } \text{H}_2\text{O} = 2.50 \text{ L of } \text{C}_3\text{H}_8 \times 4 = 10.0 \text{ L}
\][/tex]
Therefore, [tex]\(2.50\)[/tex] liters of propane ([tex]\(\text{C}_3\text{H}_8\)[/tex]) consumes [tex]\(12.5\)[/tex] liters of oxygen ([tex]\(\text{O}_2\)[/tex]) and produces [tex]\(7.5\)[/tex] liters of carbon dioxide ([tex]\(\text{CO}_2\)[/tex]) and [tex]\(10.0\)[/tex] liters of water vapor ([tex]\(\text{H}_2\text{O}\)[/tex]).
