Instead of dividing the oxygen requirement directly by the oxygen fraction of air, we can find the answer by tracking the nitrogen that comes along with the oxygen. This gives the same result through a different route.
- Set up the reaction: Methane burns as \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \), so 1 volume of methane needs 2 volumes of oxygen for complete, stoichiometric combustion, with no oxygen or methane left over.
- Find the nitrogen carried in with that oxygen: Air has oxygen and nitrogen in the ratio 21 to 79 by volume. So for every 21 volumes of oxygen, there are 79 volumes of nitrogen, giving a nitrogen to oxygen ratio of \( \frac{79}{21} = 3.762 \). For 2 volumes of oxygen, the nitrogen that comes along with it is \( 2 \times 3.762 = 7.524 \) volumes.
- Add up the whole mixture: The stoichiometric mixture is made of 1 volume methane, 2 volumes oxygen, and 7.524 volumes nitrogen. The total volume is \( 1 + 2 + 7.524 = 10.524 \).
- Work out the methane percentage: The share of methane in this total is \( \frac{1}{10.524} \times 100 = 9.5\% \), the same figure as before, now reached by tracking the inert nitrogen instead of dividing straight by the oxygen fraction.
Options 5.4% and 14.8% are the approximate lower and upper flammability limits of methane in air respectively, well known safety limits, but neither is the stoichiometric ratio. Option 10.8% does not match the oxygen and nitrogen balance worked out above, so it does not fit either.
So the stoichiometric methane concentration in the methane-air mixture is 9.5% by volume.