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This is a stoichiometry problem involving the combustion of butane \((\text{C}_4\text{H}_{10})\) with oxygen \((\text{O}_2)\) to form carbon dioxide \((\text{CO}_2)\) and water \((\text{H}_2\text{O})\). 

The balanced chemical equation for the reaction is:

\[
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
\]

From the equation:
- 2 moles of butane react with 13 moles of oxygen to produce 10 moles of water.

### Step 1: Molar masses
- Molar mass of butane \(\text{C}_4\text{H}_{10}\) = \(4 \times 12.01 + 10 \times 1.008 = 58.12 \, \text{g/mol}\)
- Molar mass of oxygen \(\text{O}_2\) = \(2 \times 16.00 = 32.00 \, \text{g/mol}\)
- Molar mass of water \(\text{H}_2\text{O}\) = \(2 \times 1.008 + 16.00 = 18.016 \, \text{g/mol}\)

### Step 2: Convert given masses to moles
- Moles of butane:  
  \[
  \text{Moles of butane} = \frac{25.0 \, \text{g}}{58.12 \, \text{g/mol}} = 0.430 \, \text{mol}
  \]
- Moles of oxygen:  
  \[
  \text{Moles of oxygen} = \frac{131 \, \text{g}}{32.00 \, \text{g/mol}} = 4.09 \, \text{mol}
  \]

### Step 3: Determine the limiting reactant
From the balanced equation, we know that 2 moles of butane react with 13 moles of oxygen. So:
- The ratio of butane to oxygen required is:
  \[
  \frac{13 \, \text{mol O}_2}{2 \, \text{mol C}_4\text{H}_{10}} = 6.5 \, \text{mol O}_2 \, \text{per} \, \text{mol C}_4\text{H}_{10}
  \]
- For 0.430 moles of butane, the required amount of oxygen is:
  \[
  0.430 \times 6.5 = 2.80 \, \text{mol O}_2
  \]
Since you have 4.09 moles of oxygen, oxygen is in excess, and butane is the limiting reactant.

### Step 4: Calculate moles of water produced
From the balanced equation, 2 moles of butane produce 10 moles of water. So, the moles of water produced are:
\[
\text{Moles of water} = 0.430 \, \text{mol C}_4\text{H}_{10} \times \frac{10 \, \text{mol H}_2\text{O}}{2 \, \text{mol C}_4\text{H}_{10}} = 2.15 \, \text{mol H}_2\text{O}
\]

### Step 5: Convert moles of water to grams
\[
\text{Mass of water} = 2.15 \, \text{mol} \times 18.016 \, \text{g/mol} = 38.8 \, \text{g}
\]

Thus, the theoretical yield of water is **38.8 g**.

Would you like any more details or have questions? Here are some follow-up questions for practice:

1. How do you determine the limiting reactant in a reaction?
2. What role does the balanced equation play in stoichiometry?
3. Why is it necessary to convert masses to moles before doing stoichiometric calculations?
4. What would happen if butane were in excess instead of oxygen?
5. How can you check if the answer has the correct number of significant digits?

**Tip:** Always ensure that the number of significant figures in your final answer reflects the precision of the given data.

What is the theoretical yield of water formed from the reaction of 25.0 g of butane and 131.0 g of oxygen gas?

Learn how to calculate the theoretical yield of water from the combustion of butane (25.0 g) with oxygen (131.0 g). This stoichiometry problem involves mass-to-mole conversions and identifying the limiting reactant.

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