Problem 85 - Chemistry The Central Science In Si Units

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 \(S_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2},\) how many grams of \(S_{8}\) would be produced, assuming \(98 \%\) yield?

Step-by-Step Solution

Answer

27.66 grams of \(\text{S}_8\) will be produced.

Step 1: Calculate Molar Masses

First, we calculate the molar masses of the reactants and products. Hydrogen sulfide (\(\text{H}_2\text{S}\)) has 2 hydrogens and 1 sulfur, so its molar mass is \(2 \times 1.01 + 32.07 = 34.09 \text{ g/mol}\). For oxygen (\(\text{O}_2\)), the molar mass is \(2 \times 16.00 = 32.00 \text{ g/mol}\). The sulfur octamer (\(\text{S}_8\)) has a molar mass of \(8 \times 32.07 = 256.56 \text{ g/mol}\).

Step 2: Determine Moles of Reactants

Next, calculate the number of moles of each reactant. For \(30.0 \text{ g of } \text{H}_2\text{S}\), the moles are \(\frac{30.0}{34.09} \approx 0.880 \text{ moles}\). For \(50.0 \text{ g of } \text{O}_2\), the moles are \(\frac{50.0}{32.00} \approx 1.563 \text{ moles}\).

Step 3: Identify Limiting Reactant

According to the balanced equation, 8 moles of \(\text{H}_2\text{S}\) react with 4 moles of \(\text{O}_2\). Thus, 0.880 moles of \(\text{H}_2\text{S}\) would require \(0.880 \times \frac{4}{8} = 0.440 \text{ moles of } \text{O}_2\). Since we have \(1.563 \text{ moles of } \text{O}_2\), \(\text{H}_2\text{S}\) is the limiting reactant.

Step 4: Calculate Theoretical Yield of \(\text{S}_8\)

From the stoichiometry of the reaction, 8 moles of \(\text{H}_2\text{S}\) produce 1 mole of \(\text{S}_8\). Therefore, 0.880 moles of \(\text{H}_2\text{S}\) will produce \(0.880 \times \frac{1}{8} = 0.110 \text{ moles of } \text{S}_8\).

Step 5: Convert Theoretical Yield to Grams

Convert the moles of \(\text{S}_8\) to grams using its molar mass: \(0.110 \text{ moles} \times 256.56 \text{ g/mol} = 28.22 \text{ grams of } \text{S}_8\).

Step 6: Calculate Actual Yield

Considering a 98% yield, the actual yield is \(28.22 \text{ grams} \times 0.98 = 27.66 \text{ grams of } \text{S}_8\).

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first. Once it is used up, the reaction stops, and no more product can form. Identifying the limiting reactant is crucial, as it determines the maximum amount of product that can be produced.
To find the limiting reactant, compare the mole ratio of the reactants with the ratio required by the balanced chemical equation.

• First, calculate the moles of each reactant.
• Next, use the balanced equation to see which reactant is required in the least amount to react completely with the other reactants.

In our Claus process example, rom the balanced equation, we know 8 moles of ext{H}_2 ext{S} need 4 moles of ext{O}_2. Given the available moles of each reactant, ext{H}_2 ext{S} turns out to be the limiting reactant. This is because less ext{H}_2 ext{S} is available compared to its required amount to completely react with ext{O}_2. Determining the limiting reactant ensures accurate calculations of the theoretical yield of the product.

Molar Mass

The molar mass of a substance is the mass of one mole of its particles, usually expressed in grams per mole (g/mol). It is a fundamental concept in chemistry that allows us to convert between the mass of a substance and the number of moles.
To calculate the molar mass, sum the atomic masses of all the atoms in a chemical formula, using the periodic table as a reference.

• For ext{H}_2 ext{S}, the molar mass calculation involves 2 hydrogens (each with a mass of approximately 1.01 g/mol) and 1 sulfur (32.07 g/mol), giving a total of 34.09 g/mol.
• For ext{O}_2, each oxygen has a mass of about 16.00 g/mol, leading to a molar mass of 32.00 g/mol.
• Finally, ext{S}_8, which consists of 8 sulfur atoms, has a molar mass of 256.56 g/mol.

Molar mass is crucial for stoichiometry calculations, as it directly relates mass and moles, enabling the accurate determination of products and reactants in chemical reactions.

Stoichiometry

Stoichiometry involves the quantitative analysis of reactants and products in a chemical reaction. It is based on the balanced chemical equation, which ensures that the number of atoms for each element is conserved. This balance helps us understand the mole ratio between substances involved in the reaction.
For the Claus process reaction, stoichiometry tells us that:

• 8 moles of ext{H}_2 ext{S} react with 4 moles of ext{O}_2
• This produces 1 mole of ext{S}_8 and 8 moles of water ( ext{H}_2 ext{O})

To determine the amount of product formed, utilize these mole ratios. By finding the moles of limiting reactant and applying the stoichiometric coefficients, you can calculate the theoretical yield of the product. This method enables precise predictions of the quantities of products and unconsumed reactants, optimizing reactions in both industrial and laboratory settings.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. Each reaction is represented by a chemical equation that describes this change.
In a chemical equation:

• Reactants are written on the left side
• Products on the right side
• An arrow indicates the direction of the reaction

For example, in the Claus process: 8 ext{H}_2 ext{S}(g) + 4 ext{O}_2(g) → ext{S}_8(l) + 8 ext{H}_2 ext{O}(g) This shows how hydrogen sulfide ( ext{H}_2 ext{S}) and oxygen ( ext{O}_2) combine to form sulfur ( ext{S}_8) and water ( ext{H}_2 ext{O}).

• The coefficients in the equation indicate the ratio of moles of each substance involved.
• The state symbols (g for gas, l for liquid) describe the physical state.

Understanding chemical reactions allows us to predict how substances combine, their reaction conditions, and the products formed, which is foundational in chemical analysis and industry.