Question

The volume of oxygen necessary for the complete combustion of 20 l of propane is

Answer 1

Answer:

hey mate........

If the 20 ml of propane is measured at STP, then: 

1.0 mole of gas at STP has a volume of 22.4 liters = 22400 ml 

20 ml/22400ml = 8.93 x 10^-4 moles of propane. 

Propane is C3H8, so the equation is: 

C3H8 + 5 02 ======> 3 CO2 + 4 H2O 

so, you would need 5 * 8.93x10^-4 moles = 4.46 x 10^-3 moles of oxygen gas, which, at STP, would occupy: 

22400 ml/mole * 4.46x10^-3 moles = 100 ml 

An easier way to calculate the same answer would be to realize that, from the equation: 

C3H8 + 5 O2 ======> 3 CO2 + 4 H2O 

pls mark brainlist answer

Explanation:

Answer 2

Answer:

100 L

as

5 L of oxygen is used for combustion of 1 L propane