# Stoichiometric Calculation of Coal Burning Process

Generally, coal is composed of several important chemical elements. They are carbon (C), hydrogen (H), sulfur (S), oxygen (O), and some other elements. These elements interconnect chemically to form new hydrocarbon compounds. The hydrocarbon chemical bonds stored energy. When the bond is cut off through the combustion process, the stored energy will be released into the environment. If we write into a chemical reaction, then coal combustion will look like the below:

Coal + O2 → Product + Heat Energy

The carbon burning process, produced the most heat energy for the entire process. Carbon burning also allows the formation of carbon monoxide if incomplete combustion occur. The content of hydrogen and sulfur in coal also contributes a small part of the heat energy when the combustion process takes place.

To facilitate our understanding, let us consider the following example calculations.

For example, a coal content analysis mention that the coal has the following composition:

• Carbon: 73%
• Hydrogen: 4.5%
• Oxygen: 5.9%
• Nitrogen: 1.5%
• Sulfur: 5%
• Water: 2.1%
• Ash: 8%

From the above data, we can determine the mol/100 gram value of each coal component, as well as we specify the mol/mol of carbon to be able to determine the chemical structure of the coal molecule.

So the chemical formula of coal molecule is:

CH0,74O0,061N0,018S0,026

With a slight rounding up the air composition, which is 79% nitrogen and 21% oxygen, so for one mole of oxygen, there are 3,762 nitrogen. With that data let us make a perfect burning stoichiometry reaction from the related coal.

CH0,74O0,061N0,018S0,026 + 1,211 (O2 + 3,762N2) → CO2 + 0,37H2O + 0,026SO2 + 4,565N2

From the equation of the perfect combustion above, then we can determine the air / fuel ratio, so we know how much air is needed to burn 1 kg of coal.

AFR = $\dfrac {1,211\left( 32+3,762\times 28\right)}{12+1\times 0,74+16\times 0,061+14\times 0,018+32\times 0,026}$

AFR = 11,237