What is volume of O2 required at 0oC and 1 atm to burn completely 1L propane gas

Asked by Anil | 13th May, 2017, 12:46: PM

Expert Answer:

C3H8 + 5O2 —> 3CO2 + 4H2O

22.4L    5X 22.4L

Since 22.4 L C3H8 at STP = 5x 22.4 L of O2 at STP

Hence 1 L C3H8 at STP = 5 x 22.4/22.4 of O2 at STP

                                        = 5 L of O2 at NTP

Answered by Vaibhav Chavan | 13th May, 2017, 06:26: PM