Also, what is the maximum fuel:air ratio possible in a combustion chamber of a jet engine?
Edit: to honor my original answer, I shall keep it as it is. The question was edited so much I think it would have constituted a whole new question. In it's original form, the Q presented jet engine f/a ratio as 1/4 - 1/3 and compared it to the stoich ratio of 1/15.
Your numbers seem way off, except for the stoichiometric fuel/air ratio (for kerosene).
Please see the following NASA article: Beginner's Guide to Propulsion and EngineSim Fuel and Air Relationships Activity
A quote from there:
For jet engine combustion, the important parameter is the fuel/air ratio (f/a) which is the ratio of the mass of fuel to the mass of air being burned in the engine.
The stoichiometric ratio of 1/15 can also be written as approximately 0.067
The article further elaborates on the subject:
For example, at sea level an airliner taking off at 375 mph with the throttle at 100% has an f/a of .017
This 0.017 translates into about 1/59
No matter how you look at the ratio, mass or volume, in no case will a jet engine run at f/a of 1/4~1/3. The mixture will simply be too rich to ignite and burn.
Here is a link to a Wikipedia graph about the range of typical jet engine air:fuel ratios:
Please note the graph displays the ratio in reverse order from what was used before in this answer, ranging from ~25/1 to ~180/1.
We have to remember air is only 20% oxygen. When you write the stoichimetric formula for the chemical reaction (using oxygen) on a molar basis, you get the combustion ratio by converting the fuel and oxidizer from moles to mass.
To get the combustion ratio using air, you divide the oxygen mass by its fractional proportion in air.
Interestingly, the stoichimetric fuel/oxygen ratio (by mass) is around 1:3. 3 divided by 0.21 gives us around 15 for kerosene and air.
Turbojets burn fuel in combustion "cans" within the flammable fuel/air percentage limits and use the heat from that combustion to warm more air passing through the the core.
Bypass fans can make these ratios even higher. For example, the Ge 90 has a bypass ratio of 9 to 1. Of the 10% of the total air mass flow going through the core, its air to fuel ratio is around 33 to 1, much higher than stoichiometric.
The total air to fuel ratio (by mass) is 330 to 1!
One can see why fuel combustion "cans" were invented, and why fan jets are becoming more popular.