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The given problem involves the combustion of octane (\(C_8H_{18}\)) reacting with oxygen gas (\(O_2\)) to produce carbon dioxide (\(CO_2\)) and water (\(H_2O\)). You're asked to calculate the percent yield of water based on the following information:

- 5.71 g of octane is reacted with 37.2 g of oxygen.
- 3.89 g of water is produced.
  
To solve this, we follow these steps:

### Step 1: Write the balanced chemical equation.
The combustion of octane is:

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

### Step 2: Find the molar masses.
- Molar mass of octane (\(C_8H_{18}\)):  
  \(8 \times 12.01 \, \text{g/mol} + 18 \times 1.008 \, \text{g/mol} = 114.23 \, \text{g/mol}\)

- Molar mass of oxygen (\(O_2\)):  
  \(2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}\)

- Molar mass of water (\(H_2O\)):  
  \(2 \times 1.008 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.016 \, \text{g/mol}\)

### Step 3: Determine the limiting reactant.
To find the limiting reactant, we convert the masses of octane and oxygen into moles:

- Moles of octane:
\[
\frac{5.71 \, \text{g}}{114.23 \, \text{g/mol}} = 0.04998 \, \text{mol}
\]

- Moles of oxygen:
\[
\frac{37.2 \, \text{g}}{32.00 \, \text{g/mol}} = 1.1625 \, \text{mol}
\]

From the balanced equation, \(2 \, \text{mol}\) of octane reacts with \(25 \, \text{mol}\) of oxygen. The ratio of octane to oxygen needed is:

\[
\frac{2}{25} = 0.08
\]

Now, calculate the actual ratio:
\[
\frac{0.04998}{1.1625} = 0.043
\]

Since 0.043 is less than 0.08, **oxygen is in excess**, and **octane is the limiting reactant**.

### Step 4: Calculate the theoretical yield of water.
From the balanced equation, \(2 \, \text{mol}\) of octane produces \(18 \, \text{mol}\) of water. Therefore, the molar ratio between octane and water is:

\[
\frac{18}{2} = 9
\]

Thus, 0.04998 mol of octane should produce:

\[
0.04998 \, \text{mol} \times 9 = 0.4498 \, \text{mol of water}
\]

Theoretical mass of water:
\[
0.4498 \, \text{mol} \times 18.016 \, \text{g/mol} = 8.1 \, \text{g}
\]

### Step 5: Calculate the percent yield.
The actual yield is given as 3.89 g. Thus, the percent yield is:

\[
\frac{3.89 \, \text{g}}{8.1 \, \text{g}} \times 100 = 48.0\%
\]

### Final Answer:
The percent yield of water is **48.0%**.

---

Would you like any further details on the solution or have any questions?

Here are some related questions you might find interesting:
1. How do we determine the limiting reactant in a reaction with multiple reactants?
2. Why is balancing chemical equations important in stoichiometry?
3. What factors affect the percent yield in a chemical reaction?
4. How can we improve the percent yield in a laboratory experiment?
5. How does the concept of limiting reactants apply in industrial chemical reactions?

**Tip:** Always ensure that your chemical equations are balanced before performing stoichiometric calculations.

Liquid octane (C8H18) reacts with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). If 3.89 g of water is produced from the reaction of 5.71 g of octane and 37.2 g of oxygen gas, calculate the percent yield of water.

Learn how to calculate the percent yield of water in the combustion of octane, using stoichiometry to find the limiting reactant and determine the theoretical yield. Suitable for advanced high school chemistry students.

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