Problem 85 - Chemistry: The Central Science in SI Units, Global Edition

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

The actual mass of S8 produced in this reaction, assuming a 98% yield, is \(27.6 \, grams\).

Step 1: Convert masses to moles

First, we need to convert the masses of H2S and O2 to moles. We use their respective molar masses for this conversion. Molar mass of H2S: \(1 (H) * 2 + 32 (S) = 2 + 32 = 34 g/mol\) Molar mass of O2: \(16 (O) * 2 = 32 g/mol\) Now, we can convert the masses of H2S and O2 to moles: Moles of H2S = \(\dfrac{30.0 g}{34 g/mol} = 0.882 moles\) Moles of O2 = \(\dfrac{50.0 g}{32 g/mol} = 1.563 moles\)

Step 2: Identify the limiting reactant

Next, we need to identify the limiting reactant. For this, we will use the mole ratios provided by the balanced equation: \(8 \, H_2S + 4 \, O_2 \rightarrow S_8 + 8 \, H_2O\) Divide moles of each reactant by their respective stoichiometric coefficients: Moles ratio of H2S: \(\dfrac{0.882}{8} = 0.110\) Moles ratio of O2: \(\dfrac{1.563}{4} = 0.391\) Since H2S has the lowest mole ratio, it will be the limiting reactant.

Step 3: Calculate the theoretical yield of S8

Now, we need to calculate the theoretical yield of S8 in moles, using the balanced equation. In this reaction, 8 moles of H2S produce 1 mole of S8. Moles of S8 = moles of H₂S * \(\dfrac{1 \, mole \, S_{8}}{8 \, moles \, H_{2}S}\) = \(0.882 * \dfrac{1}{8} = 0.110\) moles Now, let's calculate the theoretical yield in grams. The molar mass of S8 is: \(32 (S) * 8 = 256 g/mol\). Theoretical yield of S8 = \(0.110 \, moles \times 256 \dfrac{g}{mol}\) = \(28.16 g\)

Step 4: Apply 98% yield to find actual mass of S8 produced

Finally, we need to apply the given 98% yield to find the actual mass of S8 produced: Actual mass of S8 = theoretical yield of S8 * \% yield = \(28.16 g * 0.98 = 27.6 g\) Therefore, the actual mass of S8 produced in this reaction is 27.6 grams.

Key Concepts

Limiting Reactant

The concept of the limiting reactant is pivotal in chemical reactions. In any reaction, reactants are consumed to yield products. However, they may not always be present in stoichiometrically balanced proportions. The limiting reactant is the one that will be entirely consumed first, preventing any further reaction from occurring. In other words, it 'limits' the amount of product that can be formed. For the Claus process detailed in this exercise, determining the limiting reactant was crucial to predict the maximum amount of sulfur (\(S_8\)) that could be produced.
First, you convert the given masses of reactants into moles by dividing their mass by their molar mass. Given ratios from the balanced equation, you divide the moles of each reactant by their stoichiometric coefficients. The reactant with the smallest resulting value is your limiting reactant. In our case, it was hydrogen sulfide (\(H_2S\)), as its normalized mole value was lower than that of oxygen (\(O_2\)). Recognizing the limiting reactant is essential for accurately determining the theoretical yield of the reaction.

Molar Mass

Molar mass is an essential concept in stoichiometry that refers to the mass of a given substance (in grams) per mole of said substance. It aids in converting between mass and moles, linking macroscopic and molecular scales.
For any atomic element, the molar mass is numerically equal to its atomic weight on the periodic table and is expressed in grams per mole (g/mol). For compounds, you simply sum up the molar masses of the constituent elements according to the chemical formula. For example:

• The molar mass of hydrogen sulfide (\(H_2S\)) is calculated by adding twice the molar mass of hydrogen (2 g/mol) to the molar mass of sulfur (32 g/mol), resulting in 34 g/mol.
• The molar mass of oxygen molecule (\(O_2\)) consists of two oxygen atoms, hence 32 g/mol.

Using these molar masses, we convert given masses into moles, allowing us to further analyze and predict product formation as well as to determine the limiting reactant.

Yield Calculation

Yield calculation is a critical step in practical chemistry, highlighting how much of a product you actually obtain compared to what theoretically could be obtained. It gives insight into the efficiency of a reaction.
There are two kinds of yield: theoretical and actual. The theoretical yield is what calculations predict you could obtain, based on the limiting reactant. In our exercise, this was calculated to be 28.16 grams of (\(S_8\)).
However, the actual yield often falls short due to experimental losses or incomplete reactions. Thus, the percent yield quantifies the efficiency: \[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]
In this case, with a 98% yield stipulated, the actual mass of (\(S_8\)) was determined to be 27.6 grams. This aligns yield expectations with practical outcomes and is vital for industrial applications, ensuring resource efficiency and cost-effectiveness.