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The problem asks to determine the volume of water produced when 7.96 cm³ of methane gas (CH₄) reacts with oxygen.

### Step-by-step solution:

1. **Write the balanced chemical equation:**
   \[
   \text{CH₄} + 2\text{O₂} \rightarrow \text{CO₂} + 2\text{H₂O}
   \]
   This equation shows that 1 mole of methane (CH₄) produces 2 moles of water (H₂O).

2. **Use the ideal gas law assumption:**
   Since gases at the same conditions of temperature and pressure have the same volume per mole, the volume ratio will match the molar ratio of the balanced equation. Therefore, for every 1 volume of CH₄, 2 volumes of H₂O are produced.

3. **Calculate the volume of water:**
   Given 7.96 cm³ of methane, we can use the ratio of 2:1 (since 2 volumes of water are produced for 1 volume of methane):
   \[
   \text{Volume of water} = 7.96 \, \text{cm}^3 \times 2 = 15.92 \, \text{cm}^3
   \]

### Final Answer:
The volume of water produced is **15.92 cm³**.

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Would you like further details or have any questions?

Here are five related questions to expand on the concept:
1. How does the ideal gas law help in gas volume calculations during reactions?
2. What conditions must be assumed for the gas volumes to be directly proportional to moles?
3. How would the answer change if methane were in excess and oxygen was limited?
4. What other products form when hydrocarbons like methane undergo combustion?
5. Can this method be used for reactions involving liquids or solids, and why?

**Tip:** Always ensure the chemical equation is balanced before performing any stoichiometric calculations involving volume or moles.

Methane gas and oxygen gas react to form water vapor and carbon dioxide gas. What volume of water would be produced by this reaction if 7.96 cm³ of methane were consumed?

Learn how to calculate the volume of water produced from the combustion of 7.96 cm³ of methane gas using stoichiometry and the ideal gas law. This problem explores gas laws and volume ratios, suitable for high school students in Grades 9-12.

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