Problem 57 - Chemistry and Chemical Reactivity

Question

Oxygen and Sulfur In the "contact process" for making sulfuric acid, sulfur is first burned to \(\mathrm{SO}_{2}\). Environmental restrictions allow no more than \(0.30 \%\) of this \(\mathrm{SO}_{2}\) to be vented to the atmosphere. (a) If enough sulfur is burned in a plant to produce \(1.80 \times 10^{6} \mathrm{kg}\) of pure, anhydrous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per day, what is the maximum amount of \(\mathrm{SO}_{2}\) that is allowed to be exhausted to the atmosphere? (b) One way to prevent any \(\mathrm{SO}_{2}\) from reaching the atmosphere is to "scrub" the exhaust gases with slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2}:\) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CaSO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) \(2 \mathrm{CaSO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CaSO}_{4}(\mathrm{s})\) What mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (in kilograms) is needed to remove the \(\mathrm{SO}_{2}\) calculated in part (a)?

Step-by-Step Solution

Answer

The maximum exhaustible SO2 is 3528 kg, and 2653 kg of Ca(OH)2 is needed.

Step 1: Determine the Amount of Pure H2SO4

The problem states that the plant produces \(1.80 \times 10^{6}\) kg of pure \(\text{H}_2\text{SO}_4\) per day. We'll use this information to determine how much \(\text{SO}_2\) is needed for this production.

Step 2: Calculate Moles of H2SO4

The molar mass of \(\text{H}_2\text{SO}_4\) is 98.08 g/mol. Convert the mass of \(\text{H}_2\text{SO}_4\) produced daily into moles:\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{1.80 \times 10^{6} \text{ kg} \times 10^3 \text{ g/kg}}{98.08 \text{ g/mol}} \approx 1.835 \times 10^7 \text{ mol} \]

Step 3: Calculate the Required SO2 for H2SO4 Production

The balanced chemical equation for converting \(\text{SO}_2\) to \(\text{H}_2\text{SO}_4\) is\[ 2\text{SO}_2 + \text{O}_2 \to 2\text{SO}_3 \to \text{H}_2\text{SO}_4 \]Since 2 moles of \(\text{SO}_2\) produce 2 moles of \(\text{H}_2\text{SO}_4\), the moles of \(\text{SO}_2\) required are equal to those of \(\text{H}_2\text{SO}_4\):\[ \text{Moles of SO}_2 = 1.835 \times 10^7 \text{ mol} \]

Step 4: Calculate Maximum Allowable SO2 Emissions

The plant can emit only 0.30% of the total \(\text{SO}_2\) produced. First, calculate the total mass of \(\text{SO}_2\) using its molar mass (64.07 g/mol):\[ \text{Mass of SO}_2 = 1.835 \times 10^7 \text{ mol} \times 64.07 \text{ g/mol} = 1.176 \times 10^9 \text{ g} \]The allowable emission is 0.30% of this:\[ \text{Allowable SO}_2 = 0.003 \times 1.176 \times 10^9 \text{ g} = 3.528 \times 10^6 \text{ g} \approx 3528 \text{ kg} \]

Step 5: Chemical Reaction with Ca(OH)2

The reaction shows that 1 mole of \(\text{Ca(OH)}_2\) reacts with 1 mole of \(\text{SO}_2\). Thus, \(1.835 \times 10^5\) moles of \(\text{SO}_2\) need equivalent moles of \(\text{Ca(OH)}_2\).

Step 6: Calculate Mass of Ca(OH)2 Required

The molar mass of \(\text{Ca(OH)}_2\) is 74.10 g/mol. The required mass is:\[ \text{Mass of Ca(OH)}_2 = 3.578 \times 10^4 \text{ mol} \times 74.10 \text{ g/mol} = 2.653 \times 10^6 \text{ g} \approx 2653 \text{ kg} \]

Key Concepts

Contact Process

The contact process is a widely used industrial method for producing sulfuric acid. This process involves a series of steps where sulfur dioxide \\( (\text{SO}_2) \) is converted to sulfur trioxide \\( (\text{SO}_3) \) before being converted into sulfuric acid \\( (\text{H}_2\text{SO}_4) \). The steps are straightforward, but they need precision to ensure efficiency and safety.

Key stages of the contact process include:

• Sulfur is first burned to produce sulfur dioxide.
• \(\text{SO}_2\) is then oxidized to \(\text{SO}_3\) using a catalyst, typically vanadium(V) oxide.
• The \(\text{SO}_3\) is absorbed in concentrated sulfuric acid to form oleum, which is then diluted to form sulfuric acid.

This method is favored for its ability to produce high concentrations of sulfuric acid. Efficient recycling of heat and gases makes it environmentally and economically viable.

SO2 Emissions

Sulfur dioxide emissions are a significant environmental concern due to their impact on air quality and human health. In sulfuric acid manufacturing, strict regulations limit the amount of \(\text{SO}_2\) that can be emitted.

For example, if you are producing large quantities of sulfuric acid, only a small fraction (like 0.30%) of the total \(\text{SO}_2\) generated can be allowed to escape into the atmosphere.

To calculate the maximum allowable emissions:

• Determine the total amount of \(\text{SO}_2\) used in production.
• Calculate 0.30% of this total to find the permissible emissions.

Understanding these calculations is crucial for industries to stay compliant with environmental laws and to minimize pollution.

Slaked Lime Scrubbing

Slaked lime scrubbing is an effective way to remove sulfur dioxide from exhaust gases. It involves a chemical reaction where \(\text{Ca(OH)}_2\) reacts with \(\text{SO}_2\) to form calcium sulfite \(\text{(CaSO}_3\text{)}\), which is further oxidized to calcium sulfate \(\text{(CaSO}_4\text{)}\).

This process allows industries to reduce their \(\text{SO}_2\) emissions by capturing it before it reaches the atmosphere.

To find out how much slaked lime is needed:

• Calculate the moles of \(\text{SO}_2\) to be scrubbed.
• Use the 1:1 molar ratio between \(\text{Ca(OH)}_2\) and \(\text{SO}_2\) to find the required moles of slaked lime.
• Convert the moles of \(\text{Ca(OH)}_2\) to mass to determine the exact amount needed.

This method is vital for complying with environmental regulations and maintaining safe air quality standards.

Chemical Reactions in Industry

Chemical reactions are at the core of industrial processes, especially in the production of materials like sulfuric acid. These reactions must be carefully controlled to ensure both productivity and safety.

In industry:

• Reactions are often carried out on a large scale, requiring precise control of temperatures, pressures, and concentrations.
• Catalysts are frequently used to speed up reactions without being consumed in the process.
• Byproducts must be carefully managed to minimize environmental impact.

Understanding the chemistry behind these reactions helps optimize the processes, making them more efficient and eco-friendly. Whether dealing with gases, liquids, or solids, knowing how they react allows industries to innovate and improve their production lines.