Question

Limestone \(\left(\mathrm{CaCO}_{3}\right)\) is used to remove acidic pollutants from smokestack flue gases. It is heated to form lime \((\mathrm{CaO})\), which reacts with sulfur dioxide to form calcium sulfite. Assuming a \(70 . \%\) yield in the overall reaction, what mass of limestone is required to remove all the sulfur dioxide formed by the combustion of \(8.5 \times 10^{4} \mathrm{~kg}\) of coal that is 0.33 mass \% sulfur?

Short answer

1252.24 kg of limestone is required.

Step-by-step solution

Write down the relevant chemical reactions.

First, write the reactions involving limestone, lime, and sulfur dioxide:1. Decomposition of limestone:\[ \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \]2. Reaction of lime with sulfur dioxide:\[ \mathrm{CaO} + \mathrm{SO}_{2} \rightarrow \mathrm{CaSO}_{3} \]

Calculate the mass of sulfur in the coal.

Given that the coal is 0.33 mass % sulfur and the total mass of coal is \(8.5 \times 10^4 \, \text{kg}\):Calculate the mass of sulfur in the coal:\[ \text{Mass of sulfur} = 0.33 \% \times 8.5 \times 10^4 \, \text{kg} = 0.0033 \times 8.5 \times 10^4 \, \text{kg} = 280.5 \, \text{kg}\]

Convert the mass of sulfur to mass of sulfur dioxide.

The molar mass of sulfur (S) is 32 g/mol, and the molar mass of sulfur dioxide (SO\(_2\)) is 64 g/mol. The mass ratio of SO\(_2\) to S is 64/32 = 2.Thus, the mass of SO\(_2\) formed is:\[ \text{Mass of SO}_{2} = 280.5 \, \text{kg Sulfur} \times 2 = 561 \, \text{kg SO}_{2} \]

Determine the theoretical mass of CaCO\(_3\) required.

From the stoichiometry of the reaction \(\mathrm{CaO} + \mathrm{SO}_{2} \rightarrow \mathrm{CaSO}_{3}\), 1 mole of CaO reacts with 1 mole of SO\(_2\).Using the molar masses: \(M(\mathrm{SO}_{2}) = 64 \, \text{g/mol}\) and \(M(\mathrm{CaCO}_{3}) = 100 \, \text{g/mol}\).Calculate the moles of SO\(_2\):\[ \text{Moles of SO}_{2} = \frac{561 \, \text{kg}}{64 \, \text{g/mol}} = \frac{561 \times 10^3 \, \text{g}}{64 \, \text{g/mol}} = 8765.625 \, \text{mol} \]Since 1 mole of CaCO\(_3\) produces 1 mole of CaO, we need 8765.625 moles of CaCO\(_3\). Calculate the theoretical mass of CaCO\(_3\):\[ \text{Mass of CaCO}_{3} = 8765.625 \, \text{mol} \times 100 \, \text{g/mol} = 876562.5 \, \text{g} = 876.5625 \, \text{kg} \]

Account for the yield of the reaction.

Given a 70% yield, the actual mass of limestone required is higher than the theoretical value.Calculate the required mass of CaCO\(_3\) considering the yield:\[ \text{Required mass of CaCO}_{3} = \frac{876.5625 \, \text{kg}}{0.7} = 1252.2375 \, \text{kg} \]

Key concepts

stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It ensures that we can predict the amount of products formed from given quantities of reactants. In environmental chemistry, understanding stoichiometry is crucial as it helps in managing and predicting the reaction yields of pollutants removal processes from industrial emissions. For example, we use stoichiometry to calculate the amount of reactants needed to neutralize harmful gases emitted from factories. This precise calculation is vital for minimizing pollution and ensuring environmental protection.

limestone decomposition

Limestone decomposition is a key process in removing acidic pollutants from smokestack flue gases. The chemical formula for limestone is \( \text{CaCO}_3 \). When heated, it decomposes to form lime (CaO) and carbon dioxide (CO2) via the reaction: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]. This reaction forms the basis for the subsequent neutralization of sulfur dioxide (SO2) emitted during coal combustion. Limestone is an abundant and inexpensive material, making it an ideal choice for environmental remediation efforts in industrial processes.

sulfur dioxide removal

Sulfur dioxide (SO2) is a major pollutant formed during the combustion of coal containing sulfur. To remove SO2 from flue gases, lime (CaO) is used. Lime reacts with sulfur dioxide to form calcium sulfite \( \text{CaSO}_3 \): \[ \text{CaO} + \text{SO}_2 \rightarrow \text{CaSO}_3 \]. This reaction is a practical application of acid-base chemistry where the basic CaO neutralizes the acidic SO2. This process is advantageous because it helps in reducing air pollution and mitigating the environmental impact associated with coal combustion. Effective SO2 removal requires precise calculations to ensure the appropriate proportion of reactants.

reaction yield calculation

Reaction yield calculation is essential to quantify the efficiency of a chemical reaction. The theoretical yield is the amount of product that can be formed based on stoichiometry, assuming all reactants convert to products. In practice, yields are often lower than theoretical due to various factors such as incomplete reactions or side products. Yield percentage is calculated as: \[ \text{Yield (\text%)} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]. In our example, a 70% yield was assumed. Hence, the amount of limestone required for sulfur dioxide removal was calculated to be higher than the theoretical amount to account for this yield inefficiency. Proper yield calculations are essential for designing economical and efficient industrial processes.