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Question
Propane burns in air according to the following equation:
C3H8 + 5O2 → 3CO2 + 4H2O
What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen?
Solution
Given:
C3H8 + 5O2 → 3CO2 + 4H2O
Volume of air = 1000 cm3
Percentage of oxygen in air = 20%
From the given information
C3H8 + 5O2 → 3CO2 + 4H2O
1vol 5vols 3vols 4vols
According to Gay-Lussac’s law,
1 vol. of propane consumes 5 vol. of oxygen
Volume of oxygen = 1000 cm3 × 20% = 200 cm3
Therefore,
Volume of propane burnt for every 200 cm3 of oxygen
`= 1/5 xx 200 = 40 cm^3`
40 cm3 of propane is burnt.
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