Question

A coal has the following elemental (ultimate) analysis by weight: 78.4% carbon, 5.2% oxygen, 4.8% hydrogen, 1.4% nitrogen, 0.8% sulfur, and 9.4% ash. It is combusted with 25% excess air and the combustion is complete (thus CO2, H2O, and SO2 are formed due to combustion). The nitrogen from the air and the fuel does not undergo a reaction and leaves as N2 gas. Calculate the amount of air (in grams) required to combust 100 grams of coal. Calculate also the amount (in grams) of the gases produced to combust 100 grams of coal. (Molar mass of carbon, oxygen, hydrogen, nitrogen, sulfur, and air is 12 g/mole, 16 g/mole, 1 g/mole, 14 g/mole, 32 g/mole, and 28.97 g/mole, respectively.) Assume air has 79% N2 and 21% O2 by volume.

Answer 1

The stoichoimteric reactions associated with the the complete combustion of coal are :

Number of moles = Mass of given substance/

Let us consider 100 kg of the coal, therefore according to the data given, the coal contains:

• 78.4 kg of Carbon (C) or (78.4/12) = 6.533 moles of C
• 5.2 kg of Oxygen (O) or (5.2/32)= 0.1625 moles of O₂
• 4.8 kg of Hydrogen (H) or (4.8/2)= 2.4 moles of H₂
• 1.4 kg of Nitrogen (N) or (1.4/14)= 0.1 moles of N
• 0.8 kg of Sulphur (S) or (0.8/32)= 0.025 moles of S
• 9.4 kg of Ash

From the above reactions, it is clear that :

1 mole of carbon needs 1 moles of oxygen

1 mole of sulphur needs 1 mole of oxygen

1 mole of Hydrogen needs 0.5 moles of oxygen

Therefore for the given problem,

6.533 moles of carbon needs 6.533 moles of oxygen

2.4 moles of hydrogen needs 1.2 moles of oxygen

0.025 moles of sulphur needs 0.025 moles of oxygen

Nitrogen does not react with the oxygen and the oxygen present in coal will participate in the combustion.

Total moles needed = 6.533 + 1.2 + 0.025 = 7.758 moles of O₂

Moles of O₂ already present = 0.1625 moles of O₂

As 25% excess oxygen is needed , oxygen needed= 1.25 x 7.758 = 9.697 moles

as 0.1625 moles of O₂ is present, Moles needed =9.697 - 0.1625 = 9.535 moles

1 mole of air contains 0.79 moles of N₂ and 0.21 moles of O₂.

Therefore x moles of air contains 9.535 moles of O₂

x = (9.535 x 0.79) / 0.29 = 25.97 moles of air

therefore mass of air needed = 25.97 x 28.97 = 752.485 g

Mass of CO₂ produced = 6.533 x 12 = 78.396 g

Mass of SO₂ produced = 0.025 x 32 = 0.8 g

Mass of H₂O produced = 2.4 x 18 = 43.2 g