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Answer:
To determine the mass of \( \text{CO}_2 \) formed during the combustion of \( \text{C}_3\text{H}_8 \) (propane), follow these steps:
1. **Write the balanced chemical equation for the combustion of propane:**
The combustion of propane (\( \text{C}_3\text{H}_8 \)) in the presence of oxygen (\( \text{O}_2 \)) produces carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
\[
\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
To balance the equation:
- Carbon atoms: There are 3 carbons in \( \text{C}_3\text{H}_8 \), so we need 3 \( \text{CO}_2 \) molecules.
- Hydrogen atoms: There are 8 hydrogens in \( \text{C}_3\text{H}_8 \), so we need 4 \( \text{H}_2\text{O} \) molecules.
- Oxygen atoms: On the right side, we have \( 3 \times 2 = 6 \) oxygens from \( \text{CO}_2 \) and \( 4 \times 1 = 4 \) oxygens from \( \text{H}_2\text{O} \), totaling 10 oxygens. Thus, we need 5 \( \text{O}_2 \) molecules on the left side.
The balanced equation is:
\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\]
2. **Determine the molar masses:**
- Molar mass of \( \text{C}_3\text{H}_8 \):
\[
3 \times 12.01 + 8 \times 1.008 = 36.03 + 8.064 = 44.094 \, \text{g/mol}
\]
- Molar mass of \( \text{CO}_2 \):
\[
12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01 \, \text{g/mol}
\]
3. **Use stoichiometry to find the mass of \( \text{CO}_2 \) formed:**
According to the balanced equation, 1 mole of \( \text{C}_3\text{H}_8 \) produces 3 moles of \( \text{CO}_2 \).
If you have \( m \) grams of \( \text{C}_3\text{H}_8 \):
- Moles of \( \text{C}_3\text{H}_8 \):
\[
\text{moles of } \text{C}_3\text{H}_8 = \frac{m}{44.094}
\]
- Moles of \( \text{CO}_2 \) produced:
\[
\text{moles of } \text{CO}_2 = 3 \times \frac{m}{44.094}
\]
- Mass of \( \text{CO}_2 \) produced:
\[
\text{mass of } \text{CO}_2 = \text{moles of } \text{CO}_2 \times 44.01
\]
\[
\text{mass of } \text{CO}_2 = 3 \times \frac{m}{44.094} \times 44.01
\]
\[
\text{mass of } \text{CO}_2 = 3 \times \frac{44.01}{44.094} \times m
\]
Simplify the term:
\[
\text{mass of } \text{CO}_2 \approx 3 \times 0.997 \times m
\]
\[
\text{mass of } \text{CO}_2 \approx 2.99 \times m
\]
So, for every gram of \( \text{C}_3\text{H}_8 \) combusted, approximately \( 2.99 \) grams of \( \text{CO}_2 \) are formed.
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