Question

13.8 Coal with the mass analysis \(77.54 \%\) C, \(4.28 \%\) H, \(1.46 \% \mathrm{~S}\), \(7.72 \% \mathrm{O}, 1.34 \% \mathrm{~N}, 7.66 \%\) noncombustible ash burns completely with \(120 \%\) of theoretical air. Determine (a) the balanced reaction equation. (b) the amount of \(\mathrm{SO}_{2}\) produced, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of coal.

Short answer

Balanced reaction and 0.0295 kg SO2 per kg coal

Step-by-step solution

- Understand the Given Data

Analyze the given compositions of coal: carbon (C) - 77.54%, hydrogen (H) - 4.28%, sulfur (S) - 1.46%, oxygen (O) - 7.72%, nitrogen (N) - 1.34%, and noncombustible ash - 7.66%. The sum of these percentages should be 100% for mass balance consistency.

- Write the Theoretical Combustion Reaction

The theoretical combustion reaction for coal with composition C, H, S, O, and N involves burning with air to form CO2, H2O, SO2, N2, and any other products based on the chemical percentages. The basic form of the reaction for complete combustion is: C_xH_yS_zN_wO_u + a(O2 + 3.76N2) -> bCO2 + cH2O + dSO2 + eN2

- Determine the Stoichiometric Coefficients

Using the mass percentages, calculate the moles of each component based on 1 kg of coal and apply stoichiometry: - Moles of C: 0.7754 kg / 12.01 kg/kmol - Moles of H: 0.0428 kg / 1.008 kg/kmol - Moles of S: 0.0146 kg / 32.07 kg/kmol - Moles of O: 0.0772 kg / 16.00 kg/kmol - Moles of N: 0.0134 kg / 14.01 kg/kmol

- Calculate Theoretical Air Requirement

Calculate the air required using stoichiometry, considering the moles of O2 needed for each component: - C: Reacts with O2 to form CO2 - H: Reacts with O2 to form H2O - S: Reacts with O2 to form SO2 Use the resulting molar amounts to determine air needed for complete combustion.

- Adjust for Excess Air

Given that 120% of theoretical air is used, determine the actual air supplied. The amount of air = 1.2 * Theoretical air Translate this into the moles needed and adjust for the excess air in the balanced equation.

- Balanced Reaction Equation

Combine the information to form a complete balanced equation. Remember to account for the excess air as well.

- Calculate SO2 Produced

Using the moles of sulfur and stoichiometry, calculate the moles of SO2 produced per mole of coal combusted, then convert this to kg of SO2 per kg of coal: SO2 mass = moles of S * molecular weight of SO2 (64.07 kg/kmol) * (1.46 / 100).

Key concepts

Stoichiometry

Stoichiometry is the heart of chemical calculations. It's all about finding the right proportions. In the combustion reaction of coal, each element reacts with oxygen in specific ratios to form products. For our given coal, which mainly contains carbon, hydrogen, sulfur, oxygen, and nitrogen, we need to figure out how many moles of each substance we have. Here's a friendly reminder to always start with the mass percentages given in the problem. Next, convert these mass percentages into moles using the molecular weights of each element.
Once we have the moles, we can use these to determine the theoretical amount of oxygen needed. This involves knowing how many moles of oxygen each component will react with. For example, carbon reacts in a 1:1 ratio with oxygen to form carbon dioxide, while hydrogen needs half as much oxygen to form water. With these ratios figured out, we can predict the amounts of each product formed.

Theoretical Air

The concept of theoretical air is crucial in combustion reactions. It represents the exact amount of oxygen needed for the complete combustion of a fuel. In our problem, we need to calculate the theoretical air needed to burn the given coal composition completely. This involves calculating the moles of oxygen required for each component.
For our coal, we can summarize the oxygen requirements for complete combustion:

• Carbon (C) reacts with oxygen (O2) to form carbon dioxide (CO2).
• Hydrogen (H2) burns to form water (H2O).
• Sulfur (S) burns to form sulfur dioxide (SO2).

Each step involves multiplying the moles carried out in the stoichiometry steps by the moles of oxygen each reacts with. Then, we sum these to find the total oxygen needed. Next, we account for the additional theoretical air supplied, which in this case is 120% of the theoretical amount calculated.

Mass Balance

The mass balance of a chemical reaction ensures that mass is conserved. This principle states that the total mass of reactants must equal the total mass of products. In the combustion reaction of coal, we need to ensure that the mass of our initial elements equals the mass of the products formed plus any excess air.
To perform a mass balance in our problem, we start with 1 kg of coal and break down its composition into the respective masses. For each component, convert the mass to moles and use these to predict the mass of the products. Mass balance also ensures that the sum of all percentages in the composition of coal equals 100%. Any discrepancy means an error in composition or calculations. Therefore, our calculated masses of the products, including CO2, H2O, SO2, and N2, must mirror the total mass of coal and air inputs, adjusted for theoretical air.

Complete Combustion

Complete combustion occurs when a fuel reacts with sufficient oxygen, producing mostly carbon dioxide (CO2) and water (H2O). Incomplete combustion, on the other hand, results in products like carbon monoxide (CO) due to insufficient oxygen.
For coal combustion, complete combustion implies that carbon is fully converted to CO2 and hydrogen to H2O. Sulfur becomes SO2 in this scenario. The problem ensures that 120% of theoretical air is provided, which supports complete combustion. The balanced equation should account for all components. By following the stoichiometric coefficients and ensuring enough oxygen, we confirm complete combustion occurs. Also, excess theoretical air helps avoid incomplete combustion, ensuring all carbon and hydrogen in coal are fully oxidized. No unburned residues or undesired products remain in this reaction, ensuring efficiency and environmental compliance.