Question

) Explain step wise procedure to balance the chemical equation
for combustion of propane.​

Answer 1

Answer:

Propane – LPG – burns within its limits of flammability.

The lower and upper limits of flammability are the percentages of LPG that must be present in an LPG/air mixture.

This means that between 2.15% and 9.6% of the total LPG/air mixture must be LPG in order for it to be combustible.

However, the optimal mixture is 4% LPG/air.

So, 4 parts LPG (propane) to 96 parts air.

With complete combustion of propane, the burner produces a blue flame.

So, richer mixtures, those closer to 9.6%, are likely to suffer from incomplete combustion.

A yellow flame, soot and excessive condensation are three physical signs of incomplete combustion.

The natural gas – methane – limits of flammability are different, at 5.4% to 17%.

The optimal combustion mixture for methane is also different, at approximately 10.42%.

Complete Combustion of Propane Equation - Combustion Formula for LPG

In the complete combustion of propane equation, in the presence of enough oxygen, propane burns to form water vapour and carbon dioxide, as well as heat. So, this is the complete combustion of propane equation in both words and chemical formulas:

Propane + Oxygen → Carbon Dioxide + Water + Heat

C3H8 + 5 O2 → 3 CO2 + 4 H2O + Heat

Equation for Incomplete Combustion of Propane-LPG

The equation for incomplete combustion of propane is: 2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat.

If not enough oxygen is present for complete combustion, incomplete combustion occurs.

The result of incomplete combustion is, once again, water vapour, carbon dioxide and heat.

But it also produces carbon monoxide.

Equation for Incomplete Combustion of Propane-LPG in Words & Symbols

Gas + Oxygen = Water + Carbon Dioxide + Carbon Monoxide + Heat

2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat