Mole ConceptMedium
Question
Air contains 20% O2 by volume. What volume of air is needed at 0o C and 1 atm for complete combustion of 80 g methane?
Options
A.10 L
B.50 L
C.224 L
D.1120 L
Solution
CH4 + 2O2 → CO2 + 2 H2O
16 g 2 × 22.4 L at 0o C and 1 atm
$\therefore 80\text{ g} \frac{2 \times 22.4}{16} \times 80 = 224L$
Hence, the volume of air needed $= \frac{100}{20} \times 224 = 1120L$
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