Question

13.11 A sample of dried Appanoose County coal has a mass analysis of \(71.1 \%\) carbon, \(5.1 \%\) hydrogen \(\left(\mathrm{H}_{2}\right), 9.0 \%\) oxygen 13.11 A sample of dried Appanoose County coal has a mass analysis of \(71.1 \%\) carbon, \(5.1 \%\) hydrogen \(\left(\mathrm{H}_{2}\right), 9.0 \%\) oxygen \(\left(\mathrm{O}_{2}\right), 1.4 \%\) nitrogen \(\left(\mathrm{N}_{2}\right), 5.8 \%\) sulfur, and the rest noncombustile ash. For complete combustion with the theoretical amount of air, determine (a) the amount of \(\mathrm{SO}_{2}\) produced, in kg per \(\mathrm{kg}\) of coal. (b) the air-fuel ratio on a mass basis.

Short answer

a) 0.116 kg \mathrm{SO}_{2} per kg of coal. b) 10.19 kg air per kg of coal.

Step-by-step solution

- Determine the Mass of Each Element

Given that the sample is analyzed on a mass basis, the percentages can be directly translated to mass fractions. For 1 kg of coal: - Carbon: 0.711 kg - Hydrogen: 0.051 kg - Oxygen: 0.090 kg - Nitrogen: 0.014 kg - Sulfur: 0.058 kg

- Calculate the Mass of SO2 Produced

Each kg of coal contains 0.058 kg of sulfur. To find the mass of \(\mathrm{SO}_{2}\) produced, use the molar masses: \(M(\mathrm{S}) = 32 \) g/mol and \(M(\mathrm{SO}_{2}) = 64 \) g/mol. The amount of \(\mathrm{SO}_{2}\) produced can be found using \[ \mathrm{m}(\mathrm{SO}_{2}) = 0.058 \text{ kg S} \times \frac{64 \text{ g/mol }}{32 \text{ g/mol }} = 0.116 \text{ kg SO}_{2} \]

- Write the Combustion Reactions

The combustion reactions for the elements in coal are: - For Carbon: \(\mathrm{C} + \mathrm{O}_{2} \to \mathrm{CO}_{2}\) - For Hydrogen: \(2 \mathrm{H}_{2} + \mathrm{O}_{2} \to 2 \mathrm{H}_{2}\mathrm{O}\) - For Sulfur: \(\mathrm{S} + \mathrm{O}_{2} \to \mathrm{SO}_{2}\)

- Calculate the Theoretical Air Requirement

To find the air-fuel ratio, calculate the amount of air needed for complete combustion of each element. - For Carbon: \(0.711 \text{ kg C} \times \frac{32 \text{ g O}_{2}}{12 \text{ g C}} = 1.898 \text{ kg O}_{2}\) - For Hydrogen: \(0.051 \text{ kg H}_{2} \times \frac{32 \text{ g O}_{2}}{4 \text{ g H}_{2}} = 0.408 \text{ kg O}_{2}\) - For Sulfur: \(0.058 \text{ kg S} \times \frac{32 \text{ g O}_{2}}{32 \text{ g S}} = 0.058 \text{ kg O}_{2}\) - Total \(\mathrm{O}_{2} = 1.898 + 0.408 + 0.058 = 2.364 \text{ kg }\)

- Account for the Nitrogen in Air

Air contains approximately 23.2\text{%} \(\mathrm{O}_{2}\) by mass. Therefore, the mass of air needed is: \[ \text{Mass of air} = \frac{2.364 \text{ kg }}{0.232} = 10.19 \text{ kg air/kg coal} \]

- Summarize Results

Using the above calculations: - (a) The amount of \(\mathrm{SO}_{2}\) produced is 0.116 kg per kg of coal. - (b) The air-fuel ratio is 10.19 kg air per kg of coal.

Key concepts

mass analysis

Mass analysis is a crucial step in understanding the composition of a material. In this example, we look at coal, breaking down its composition by mass percentage. For 1 kg of Appanoose County coal, we have:

• 71.1% Carbon (0.711 kg)
• 5.1% Hydrogen (0.051 kg)
• 9.0% Oxygen (0.090 kg)
• 1.4% Nitrogen (0.014 kg)
• 5.8% Sulfur (0.058 kg)
• The rest as noncombustible ash and other components

Mass analysis is simpler here because percentages directly translate to mass fractions. For any material, knowing exact mass fractions helps in subsequent calculations, such as combustion modeling or pollution predictions.

air-fuel ratio

The air-fuel ratio (AFR) is key in combustion processes. It represents the mass of air needed for the complete combustion of a given mass of fuel. Let's calculate the AFR for our example coal:

• Carbon needs 1.898 kg of \(\text{O}_2\) (oxygen)
• Hydrogen needs 0.408 kg of \(\text{O}_2\)
• Sulfur needs 0.058 kg of \(\text{O}_2\)
• Total oxygen required: 2.364 kg

Knowing that air contains about 23.2% \(\text{O}_2\) by mass, we calculate the total air required:

• \(\text{Mass of air} = \frac{2.364 \text{ kg \(\text{O}_2\)}}{0.232} = 10.19 \text{ kg air/kg coal}\)

This AFR ensures complete combustion, minimizing excess fuel and optimizing pollution control.

SO2 production

Sulfur dioxide (SO2) is a significant pollutant produced during coal combustion. We calculate this using the sulfur content in coal. For every kg of coal, with 0.058 kg of sulfur:

• Molar mass of \(\text{S}\) = 32 g/mol
• Molar mass of \(\text{SO}_2\) = 64 g/mol

Using these molar masses, we find the mass of \(\text{SO}_2\) produced:

• \(\text{m}(\text{SO}_2) = 0.058 \text{ kg S} \times \frac{64 \text{ g/mol \(\text{SO}_2\)}}{32 \text{ g/mol \(\text{S}\)}} = 0.116 \text{ kg SO}_2\)

Thus, each kg of coal produces 0.116 kg of SO2. Controlling SO2 emissions is crucial for reducing air pollution and its harmful effects on health and the environment.

stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. Here, it helps us understand combustion, where coal's elements react with oxygen:

• Carbon: \(\text{C} + \text{O}_2 \to \text{CO}_2\)
• Hydrogen: \(\text{2 H}_2 + \text{O}_2 \to 2 \text{H}_2\text{O}\)
• Sulfur: \(\text{S} + \text{O}_2 \to \text{SO}_2\)

These reactions show how carbon converts to CO2, hydrogen to water, and sulfur to SO2 in the presence of oxygen. Stoichiometric calculations ensure we balance these reactions accurately, optimizing fuel use and minimizing pollutants. This is vital in industries like energy production, ensuring efficient and cleaner processes.