I would like to understand how to calculate the complete reaction of propellants at chamber temperature and pressure.
For instance taking ethyl alcohol (75%) and LOX combustion with mixture ratio of 1.4, how do we predict the complete product at the end of the reaction?
Do these in fact react completely?
Partial answer alert! I can't work your problem exactly, but this should get you started at least.
You don't say if the mixture ratio of 1.4 is mol/mol or weight/weight, but let's give this a try.
According to braeunig.us's Combustion & Exhaust Velocity
Mixture ratio is defined as the mass flow of oxidizer divided by the mass flow of fuel.
It lists the following reactions for examples. The first one is complete or stoichiometric:
CH4 + 2O2 -> CO2 + 2H2O
and the weight ratio of oxidizer to fuel is (2 x 32) / 16 or 4:1, and the second one is incomplete or non-stoichiometric:
C12H26 + 12.5 O2 -> 12CO + 13 H2O
because the carbon is only partially oxidized to carbon monoxide, not carbon dioxide. The ratio of oxidizer to fuel is (12.5 x 32) / (12 x 12 + 26 x 1) = 400/170 = 2.35.
...which is typical of many rocket engines using kerosene, or RP-1, fuel.
The optimum mixture ratio is typically that which will deliver the highest engine performance (measured by specific impulse), however in some situations a different O/F ratio results in a better overall system. For a volume-constrained vehicle with a low-density fuel such as liquid hydrogen, significant reductions in vehicle size can be achieved by shifting to a higher O/F ratio. In that case, the losses in performance are more than compensated for by the reduced fuel tankage requirement. Also consider the example of bipropellant systems using NTO/MMH, where a mixture ratio of 1.67 results in fuel and oxidizer tanks of equal size. Equal sizing simplifies tank manufacturing, system packaging, and integration.
For your example I don't know exactly what ratio "ethyl alcohol (75%)" means, so for my example answer I will do 3 moles of alcohol to 1 mole of water. If it turns out to be 75% by weight you can modify this analysis accordingly. Let's assume that the water in the fuel does not participate in the reaction, and ends up as water in the exhaust.
Propellant: 3 C2H6O + H2 (assumed 3:1 mol/mol)
Oxidizer: X O2 where we need to solve for X for "complete combustion" which basically means that all of the carbon ends up completely oxidized as CO2.
3 C2H6O + H2O + X O2 = Y CO2 + Z H2O
There are 6 carbons on the left so we know Y = 6.
There are 4 + 2X oxygens on the left and 2Y + Z on the right, so we then know 4 + 2X = 12 + Z or 2X = 8 + Z.
There are 20 hydrogens on the left and 2Z on the right, so Z = 10. That makes X = 9, and so finally:
3 C2H6O + H2O + 9 O2 = 6 CO2 + 10 H2O
(double-checking; both sides have 6C + 20H + 22O)
The ratio is (9 x 32) / (3 x (24 + 6 + 16) + 18) = 1.85 which is of course different than what you've asked for, but this should get you at least started doing stoichiometric calculations.
