What volume of oxygen at s.t.p is required to burn completely 7.5 dm ^3 of methane according to the following…

Analyze the question: We need the volume of oxygen that burns 7.5 dm³ of methane completely. Both are gases at the same conditions, so we can use Gay-Lussac's law of combining volumes.

Concept: At the same temperature and pressure, the volume ratio of gases equals the mole ratio in the balanced equation (Avogadro's law).

Balanced equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

(The question prints "20₂", which is meant to be 2O₂, and the equation as written is not fully balanced because the water needs a coefficient of 2. This does not affect the oxygen ratio.)

Solution:

• From the equation, 1 volume of CH₄ reacts with 2 volumes of O₂.
• Volume of O₂ = 2 × 7.5 dm³ = 15.0 dm³

Answer: 15.0 dm³, which is option C.

Tip: For gas-volume questions, you do not need to convert to moles or use 22.4 dm³. If all the gases are at the same conditions, just read the coefficients as volume ratios. Also, always check that the equation is balanced first.