Chemistry
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions? (assuming oxygen is 1/5th of air)
C3H8 + 5O2 ⟶ 3CO2 + 4H2O
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Answer
Given, oxygen is 1⁄5th of air = of 500 = 100 cm3
[By Gay Lussac's law]
5 Vol. of O2 requires 1 Vol. of propane
∴ 100 cm3 of O2 will require = x 100 = 20 cm3
Hence, propane burnt = 20 cm3 or 20 cc
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