Problem 144 - Chemistry An Atoms First Approach

Question

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when 120. g of methane is reacted with an equal mass of sulfur.

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction of methane and sulfur is \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\). When 120 g of methane is reacted with an equal mass of sulfur, the theoretical yield of carbon disulfide is 71.2 g.

Step 1: Write the unbalanced chemical equation

The unbalanced chemical equation is: \(CH_4 + S \rightarrow CS_2 + H_2S\)

Step 2: Balance the chemical equation

To balance the chemical equation, we need to have equal numbers of each element on both sides of the equation. Only the hydrogen atoms are not balanced now, so we should add a coefficient of 2 in front of H2S. The balanced chemical equation is: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\) b. Theoretical yield of carbon disulfide

Step 1: Calculate the number of moles of methane and sulfur

Using the molar mass of methane (16.04 g/mol) and sulfur (32.07 g/mol), calculate the number of moles of both reactants. Moles of methane: \[\frac{120 \, g}{16.04 \, g/mol} = 7.48 \, mol\] Moles of sulfur: \[\frac{120 \, g}{32.07 \, g/mol} = 3.74 \, mol\]

Step 2: Identify the limiting reactant

Compare the mole ratios of methane and sulfur in the balanced chemical equation with the mole ratios of the reactants. Ratio of methane: \[\frac{7.48 \, mol}{1} = 7.48\] Ratio of sulfur: \[\frac{3.74 \, mol}{4} = 0.935\] Since the ratio of sulfur is smaller than the ratio of methane, sulfur is the limiting reactant.

Step 3: Calculate the theoretical yield of carbon disulfide

Use the mole ratio of sulfur to carbon disulfide from the balanced chemical equation and the molar mass of CS2 (76.13 g/mol) to calculate the theoretical yield. Moles of CS2: \[\text{Moles of S} \times \frac{\text{Moles of CS2}}{\text{Moles of S}} = 3.74 \, mol \times \frac{1 \, mol \, CS2}{4 \, mol \, S} = 0.935 \, mol \, CS2\] Theoretical yield of CS2: \[0.935 \, mol \times 76.13 \, g/mol = 71.2 \, g\] Therefore, the theoretical yield of carbon disulfide is 71.2 g.

Key Concepts

Balancing Chemical Equations

The first crucial step in stoichiometry is balancing the chemical equation. This process ensures that the number of atoms for each element is the same on both sides of the equation. In the original exercise, we started with the reaction: \(CH_4 + S \rightarrow CS_2 + H_2S\). This equation isn't balanced because the number of hydrogen atoms differs on each side.
When balancing chemical equations, you follow these simple steps:

• Write down the number of each type of atom on both sides of the equation.
• Adjust coefficients to get equal numbers of atoms for each element.
• Start balancing with elements that appear in only one reactant and product, usually.
• Continue adjusting coefficients for other elements as needed.

In our example, the hydrogens were unequal. By placing a coefficient of 2 in front of \(H_2S\), we balanced the equation to \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\).
Balancing not only respects the law of conservation of mass but also provides the correct stoichiometric coefficients used in further calculations.

Limiting Reactant

Understanding the concept of the limiting reactant is essential in predicting the amount of products formed in a chemical reaction. This is the reactant that is completely consumed first, limiting the extent of the reaction. Identifying it requires comparing the mole ratios of the reactants used to those in the balanced equation.
In the given problem, we calculated the number of moles for both methane (\(CH_4\)) and sulfur (S) using their respective molar masses:

• Methane: \(\frac{120 \, g}{16.04 \, g/mol} = 7.48 \, mol\)
• Sulfur: \(\frac{120 \, g}{32.07 \, g/mol} = 3.74 \, mol\)

Next, we compared these moles based on our balanced chemical equation, \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\). Because the equation requires 4 moles of sulfur for every mole of methane, sulfur becomes the limiting reactant, as seen by its lower computed ratio of \(0.935\).
Recognizing the limiting reactant is crucial, as it dictates the maximum theoretical yield of products.

Theoretical Yield Calculation

Theoretical yield is the amount of product that could be formed in a reaction if it proceeds perfectly, without any losses. Calculating it depends on the stoichiometry of the balanced equation and the amount of the limiting reactant.
After identifying sulfur as the limiting reactant, the reaction's progress is determined by it. Using the balanced equation \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\), we calculate the moles of \(CS_2\) produced. From the balanced equation:

• We utilized the ratio: \(4 \text{ moles S} = 1 \text{ mole CS}_2\)
• Thus, \(3.74 \, mol \, S \times \frac{1 \, mol \, CS_2}{4 \, mol \, S} = 0.935 \, mol \, CS_2\)

To find the theoretical yield, we multiply the moles of the product \(CS_2\) by its molar mass, \(76.13 \, g/mol\):

• Theoretical yield: \(0.935 \, mol \times 76.13 \, g/mol = 71.2 \, g\)

This calculated figure represents the maximum amount of \(CS_2\) that can be formed from the given reactants, assuming complete reaction and no product loss.