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This problem involves a chemical reaction and asks you to determine the maximum mass of carbon dioxide (\(CO_2\)) that could be produced from the combustion of octane (\(C_8H_{18}\)) with oxygen (\(O_2\)).

### Given:
- Mass of octane = 18.0 g
- Mass of oxygen = 122.0 g

### Reaction:
The balanced combustion reaction for octane is:

\[
2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O
\]

### Steps to solve:

1. **Molar Masses**:
   - Molar mass of octane (\(C_8H_{18}\)) = \(8 \times 12.01 + 18 \times 1.008 = 114.23 \, \text{g/mol}\)
   - Molar mass of oxygen (\(O_2\)) = \(2 \times 16.00 = 32.00 \, \text{g/mol}\)
   - Molar mass of carbon dioxide (\(CO_2\)) = \(12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}\)

2. **Determine the Limiting Reactant**:
   To find the limiting reactant, we need to calculate how much \(O_2\) is needed to react with 18.0 g of octane.

   - Moles of octane:
   \[
   \text{moles of } C_8H_{18} = \frac{18.0 \, \text{g}}{114.23 \, \text{g/mol}} = 0.1575 \, \text{mol}
   \]
   According to the balanced equation, 2 moles of octane react with 25 moles of oxygen. So, for 0.1575 moles of octane:
   \[
   \text{moles of } O_2 \, \text{needed} = 0.1575 \, \text{mol} \times \frac{25}{2} = 1.96875 \, \text{mol}
   \]
   - Moles of oxygen available:
   \[
   \text{moles of } O_2 = \frac{122.0 \, \text{g}}{32.00 \, \text{g/mol}} = 3.8125 \, \text{mol}
   \]

   Since the reaction only needs 1.96875 moles of \(O_2\) and there is more \(O_2\) available (3.8125 moles), **octane** is the limiting reactant.

3. **Calculate the Maximum Mass of \(CO_2\)**:
   From the balanced equation, 2 moles of octane produce 16 moles of \(CO_2\). Therefore, for 0.1575 moles of octane:
   \[
   \text{moles of } CO_2 = 0.1575 \, \text{mol} \times \frac{16}{2} = 1.26 \, \text{mol}
   \]
   - Mass of \(CO_2\):
   \[
   \text{mass of } CO_2 = 1.26 \, \text{mol} \times 44.01 \, \text{g/mol} = 55.45 \, \text{g}
   \]

### Answer:
The maximum mass of carbon dioxide that could be produced is **55.5 g**, considering significant figures.

Would you like further details on any part of the calculation?

Here are some related questions you might be interested in:
1. How do you determine the limiting reactant in chemical reactions?
2. How does the stoichiometric ratio affect the calculation of product masses?
3. What are the general steps for performing mass-to-mass calculations in chemistry?
4. How do significant figures impact the final answer in chemical problems?
5. Why is it important to balance chemical equations before solving stoichiometry problems?

**Tip**: Always start with a balanced chemical equation to ensure the correct stoichiometric ratios are used!

Liquid octane (C₈H₁₈) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). Suppose 18.0 g of octane is mixed with 122.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Learn how to calculate the maximum mass of CO₂ produced from the combustion of 18.0 g of octane with 122.0 g of oxygen. This problem involves stoichiometry, determining the limiting reactant, and significant figures.

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