Chemistry, asked by shrutiravichandran02, 1 year ago

calculate the volume of oxygen required for the complete combustion of 8.8 grams of propane

Answers

Answered by Anonymous

3

C3H8 + 5O2 = 3CO2 + 4H2O

At NTP

44g of propane requires 5*22.4L of oxygen for complete combustion

2.2g of propane requires 2.2*5*22.4/44 = 5.6L

Volume of oxygen required for complete combustion is 5.6L

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