Question

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with 300 grams of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0\) grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{\mathrm{g}}\) would be produced, assuming \(98 \%\) yield?

Short answer

Under optimal conditions with a 98% yield, 98.0 grams of sulfur (S₈) would be produced from the given amounts of 300 grams of H₂S and 50.0 grams of O₂.

Step-by-step solution

Find the limiting reactant

First, we need to determine the amount of moles for both H₂S and O₂ using their respective molar masses. The molar mass of H₂S is approximately 34.1 g/mol, and the molar mass of O₂ is 32.0 g/mol. Moles of H₂S = (300 g) / (34.1 g/mol) = 8.80 mol Moles of O₂ = (50.0 g) / (32.0 g/mol) = 1.56 mol Now, let's compare the mole ratio of the reactants to their stoichiometric ratio in the balanced equation (8:4 or 2:1). Mole ratio = (8.80 mol H₂S) / (1.56 mol O₂) = 5.64 Since the mole ratio is greater than 2, O₂ is the limiting reactant because it is consumed faster than H₂S.

Calculate the theoretical yield of S₈

Using the stoichiometry from the balanced equation, we can calculate the theoretical yield of S₈, if all the limiting reactant (O₂) is used up. From the balanced equation, 4 moles of O₂ react to produce 1 mole of S₈. So, Moles of S₈ = (1.56 mol O₂) × (1 mol S₈ / 4 mol O₂) = 0.39 mol S₈ Now, we convert the moles of S₈ to grams. The molar mass of S₈ is approximately 256.5 g/mol. Mass of S₈ (theoretical yield) = (0.39 mol S₈) × (256.5 g/mol) = 100.0 g

Calculate the actual yield of S₈, considering the 98% yield

The actual yield of S₈ is 98% of the theoretical yield. To find this, we will multiply the theoretical yield by 0.98 (or 98%). Actual yield = 100.0 g × 0.98 = 98.0 g Thus, under optimal conditions with a 98% yield, 98.0 grams of sulfur (S₈) would be produced from the given amounts of H₂S and O₂.

Key concepts

Hydrogen Sulfide Removal

Hydrogen sulfide (H₂S) is a harmful impurity often found in natural gas. Its removal is essential to prevent corrosion and environmental damage. One of the most effective ways to eliminate H₂S from natural gas is through the Claus process. This chemical process helps convert H₂S into elemental sulfur, a less harmful material. The Claus process involves a reaction between H₂S and oxygen (O₂), resulting in the formation of sulfur (S₈) and water (H₂O). It's important to control the conditions to ensure the reaction proceeds with a high yield. This process is essential for making natural gas safe for use and minimizing the release of toxic substances into the atmosphere.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that determines the maximum amount of product that can be formed. It gets completely used up first, and once it's consumed, no more product can form, even if there are other reactants left. In the given reaction, H₂S and O₂ are the reactants. To find out which one is the limiting reactant, we calculate the moles of each based on their given masses and molar masses. In our example:

• Moles of H₂S = 8.80 mol
• Moles of O₂ = 1.56 mol

We then compare the mole ratio of the reactants with the stoichiometric ratio from the balanced equation. Since the real mole ratio was larger than the stoichiometric ratio of 2:1, O₂ is the limiting reactant. Knowing the limiting reactant is crucial as it helps in calculating the amount of product formed (in this case, sulfur S₈).

Theoretical Yield

Theoretical yield is the amount of product expected from a reaction if all of the limiting reactant is transformed completely into the product. It represents the maximum possible amount of product under perfect conditions. In the equation provided, using O₂ as the limiting reactant and the stoichiometric coefficients, we calculate the theoretical yield of sulfur (S₈). Based on the chemical equation, for every 4 moles of O₂ consumed, 1 mole of S₈ is produced. After determining that 1.56 moles of O₂ will yield 0.39 moles of S₈, we convert this to grams using the molar mass of S₈. The calculated theoretical yield of S₈ is 100.0 grams. Theoretical yield gives an understanding of the efficiency of a reaction and helps predict the amount of product.

Chemical Stoichiometry

Chemical stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products in a reaction. It allows us to calculate quantities like moles, masses, and yields. In the Claus process, stoichiometry helps understand the conversion between moles of reactants and moles of products. It requires a balanced equation, which reflects the principle of conservation of mass, indicating that matter is neither created nor destroyed. Stoichiometry allows us to determine the moles of H₂S and O₂ required and how much S₈ could be produced. It also provides the foundation for calculating the theoretical yield. By converting quantities between moles and grams through the use of molar masses, stoichiometry ensures precise chemical calculations and efficient reaction planning.