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