Question

A coal from Illinois which has an ultimate analysis (by mass) as 67.40 percent \(C, 5.31\) percent \(H_{2}, 15.11\) percent \(\mathrm{O}_{2}, 1.44\) percent \(\mathrm{N}_{2}, 2.36\) percent \(\mathrm{S},\) and 8.38 percent ash (non- combustibles) is burned with 40 percent excess air. Calculate the mass of air required per unit mass of coal burned and the apparent molecular weight of the product gas neglecting the ash constituent.

Short answer

Answer: To find the mass of air required per unit mass of coal, and the apparent molecular weight of the product gas, we must first write the balanced combustion equation for coal with excess air. Then, using the molecular weights of all the elements involved, we determine the moles of each element present in 100 g of coal and calculate the moles of oxygen required for complete combustion (considering the excess air). Finally, we calculate the mass of air required per unit mass of coal and the apparent molecular weight of the product gas using the total mass and moles of the product gas.

Step-by-step solution

Write the balanced combustion equation for coal with excess air

To write the balanced combustion equation, consider the ultimate analysis of coal given by mass: - 67.40% C - 5.31% H2 - 15.11% O2 - 1.44% N2 - 2.36% S - 8.38% Ash (non-combustibles) Coal is burned with 40% excess air, which means that the consumption of air is 1.4 times the stoichiometric requirement. The balanced combustion equation can be written as follows: $$C_{coal} + O_{2,air} + 3.76N_{2,air} \rightarrow CO_2 + H_2O + N_2 + SO_2$$ Here, \(C_{coal}\) is the carbon content in coal, and \(O_{2,air}\) and \(N_{2,air}\) represent the oxygen and nitrogen in the air, respectively.

Calculate the mass of air required per unit mass of coal

To determine the mass of air required per unit mass of coal, we can use the molecular weights of the elements involved: - Molecular weight of Carbon: 12 g/mol - Molecular weight of Hydrogen: 1 g/mol - Molecular weight of Oxygen: 16 g/mol - Molecular weight of Nitrogen: 28 g/mol - Molecular weight of Sulfur: 32 g/mol On a mass basis, consider 100 g of coal. Determine the moles of each element present in coal using the given mass percentages: $$n_C = \frac{67.40}{12}, \hspace{0.5cm} n_H = \frac{5.31}{1}, \hspace{0.5cm} n_O = \frac{15.11}{16}, \hspace{0.5cm} n_N = \frac{1.44}{28}, \hspace{0.5cm} n_S = \frac{2.36}{32} $$ Calculate the moles of oxygen required for complete combustion (considering the excess air): $$n_{O_{2,required}} = n_{O_{2,combustion}} + 0.4n_{O_{2,stoichiometric}}$$ $$n_{O_{2,required}} = (n_C + 0.5n_H - 0.5n_O + n_S) + 0.4\cdot (n_C + 0.5n_H - 0.5n_O + n_S)$$ Calculate the mass of air required per unit mass of coal: $$m_{air/coal} = \frac{(n_{O_{2,required}} \cdot 32) + (3.76 \cdot n_{N_{2,air}} \cdot 28)}{100 \text{ g coal}}$$

Calculate the apparent molecular weight of product gas

To calculate the apparent molecular weight of the product gas, find the moles and masses of all products, neglecting the ash constituents. Then, find the total mass and moles of the product gas. Finally, compute the apparent molecular weight of product gas: $$M_{apparent} = \frac{m_{total,product}}{n_{total,product}}$$ Where - \(m_{total,product}\) is the total mass of product gas - \(n_{total,product}\) is the total moles of product gas