Stoichiometry (Gaseous Volumes) — Methane Combustion Products Consider complete combustion of methane (CH4) with oxygen. At the same temperature and pressure (ideal-gas behavior), is the statement correct that 1 m^3 of methane requires 2 m^3 of oxygen and produces 1 m^3 of carbon dioxide and 2 m^3 of water vapor (steam)?

Introduction / Context:
Volume (or mole) ratios for reacting gases follow directly from balanced chemical equations when temperature and pressure are the same before and after reaction and all species are considered in the gaseous state. This is widely used in burner design and flue-gas calculations.

Given Data / Assumptions:

• Balanced reaction for complete combustion of methane.
• Ideal-gas behavior; equal T and p conditions.
• Water is considered as vapor (steam) in the products.

Concept / Approach:
The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O (vapor). Because equal gas volumes represent equal moles at the same T and p, the coefficients provide the volume ratios directly: 1 : 2 → 1 : 2.

Step-by-Step Solution:
Write stoichiometry: CH4 + 2 O2 → CO2 + 2 H2O.Match volumes to coefficients: 1 m^3 CH4 requires 2 m^3 O2.Products: 1 m^3 CO2 and 2 m^3 H2O vapor (assuming steam, not liquid water).Thus the statement is correct under the stated conditions.

Verification / Alternative check:
On a molar (or volumetric) basis, air requirement can also be inferred: 1 m^3 CH4 needs 2 m^3 O2, which corresponds to 2 / 0.21 ≈ 9.52 m^3 of air theoretically, confirming consistency.

Why Other Options Are Wrong:
“Incorrect” would only be justified if water were condensed to liquid (not a gas) or if T, p were not matched; the prompt explicitly allows water or steam, with the gaseous case implied.

Common Pitfalls:
Forgetting that the simple volume–coefficient mapping holds only for gases at the same T and p; ignoring the nitrogen in air when calculating flue-gas volumes.

Final Answer:
Correct