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The problem in the image provides a chemical reaction and asks to calculate the percent yield of water. Let’s break this down:

### Reaction Overview:
- Ethane \((C_2H_6)\) reacts with oxygen \((O_2)\) to form carbon dioxide \((CO_2)\) and water \((H_2O)\).
- The balanced equation for the combustion of ethane is:

\[
2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O
\]

### Given Data:
- **Mass of ethane**: 18.3 g
- **Mass of oxygen**: 41.7 g
- **Actual mass of water produced**: 5.03 g

### Steps to Solve:

1. **Calculate moles of ethane**:
   - Molar mass of ethane \((C_2H_6)\) = \( 2 \times 12.01 \, \text{g/mol} + 6 \times 1.008 \, \text{g/mol} = 30.07 \, \text{g/mol} \)
   - Moles of ethane = \(\frac{18.3 \, \text{g}}{30.07 \, \text{g/mol}} \approx 0.608 \, \text{mol}\)

2. **Determine moles of water produced (theoretical yield)**:
   - According to the balanced equation, for every 2 moles of ethane, 6 moles of water are produced. So, for \(0.608 \, \text{mol}\) of ethane:
   \[
   \text{Moles of water} = 0.608 \, \text{mol} \times \frac{6}{2} = 1.824 \, \text{mol}
   \]

3. **Convert moles of water to grams** (theoretical yield):
   - Molar mass of water \((H_2O)\) = \( 2 \times 1.008 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.016 \, \text{g/mol} \)
   - Mass of water = \( 1.824 \, \text{mol} \times 18.016 \, \text{g/mol} \approx 32.86 \, \text{g} \)

4. **Calculate the percent yield**:
   \[
   \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{5.03 \, \text{g}}{32.86 \, \text{g}} \times 100 \approx 15.3\%
   \]

### Final Answer:
The percent yield of water is approximately **15.3%**.

Would you like a more detailed explanation or have any questions?

### Related Questions:
1. How do you determine the limiting reactant in this reaction?
2. What is the theoretical yield of carbon dioxide in this reaction?
3. How would the percent yield change if more ethane was used?
4. How do you balance a combustion reaction like this one?
5. What other products, besides carbon dioxide and water, could form in incomplete combustion?

### Tip:
Always ensure that you use the correct molar ratios from the balanced equation when calculating theoretical yields.

Gaseous ethane (C2H6) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 5.03 g of water is produced from the reaction of 18.3 g of ethane and 41.7 g of oxygen gas, calculate the percent yield of water.

Learn how to calculate the percent yield of water in a combustion reaction involving ethane and oxygen. Given the masses of ethane (18.3 g) and oxygen (41.7 g), and the actual yield of water (5.03 g), this problem explores stoichiometry and percent yield calculations. Suitable for high school students studying chemistry.

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