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.

Short answer

The balanced equation for the reaction of methane and sulfur is: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\). When 120g of methane is reacted with an equal mass of sulfur, the theoretical yield of carbon disulfide is approximately 571.13 grams.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the unbalanced equation for the given reaction: Methane (CH4) + Sulfur (S) -> Carbon Disulfide (CS2) + Hydrogen Sulfide (H2S) Now, we can balance the equation: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\) So, the balanced equation is: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\)

Find the limiting reagent

The problem states that 120g of methane is reacted with an equal mass of sulfur. We are given the following information: - Mass of CH4 = 120g - Mass of S = 120g To identify the limiting reagent, convert the masses to moles using their respective molar masses and compare the mole ratios. Molar mass of CH4 = 12.01 (C) + 4.00 (H) = 16.01 g/mol Moles of CH4 = mass of CH4 / molar mass of CH4 = 120g / 16.01 g/mol ≈ 7.50 mol Molar mass of S = 32.07 g/mol Moles of S = mass of S / molar mass of S = 120g / 32.07 g/mol ≈ 3.74 mol Compare the mole ratios: Mole ratio of CH4 to S (from balanced equation) = 1:4 Mole ratio of CH4 to S (from given amounts) = 7.50:3.74 ≈ 2:1 Since the mole ratio for the given amounts is less than the mole ratio in the balanced equation, CH4 is the limiting reagent.

Calculate the theoretical yield of CS2

Using the balanced equation, we can find the theoretical yield of CS2 based on the moles of the limiting reagent (CH4). From the balanced equation, we see that 1 mole of CH4 produces 1 mole of CS2: 1 mol CH4 -> 1 mol CS2 We have calculated 7.50 mol of CH4, so theoretically, 7.50 mol of CH4 can react to form 7.50 mol of CS2. Now, we can convert moles of CS2 to grams using its molar mass: Molar mass of CS2 = 12.01 (C) + 2 * 32.07 (S) = 76.15 g/mol Mass of CS2 = moles of CS2 * molar mass of CS2 = 7.50 mol * 76.15 g/mol ≈ 571.13 g Theoretical yield of carbon disulfide is approximately 571.13 grams.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is like solving a puzzle where each piece has to fit perfectly. In chemistry, this puzzle ensures that the number of atoms of each element is the same on both sides of the equation. This is important because of the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

The given reaction involves methane (\(\mathrm{CH}_{4}\)) and sulfur (\(\mathrm{S}\)) producing carbon disulfide (\(\mathrm{CS}_{2}\)) and hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)). To balance it, you need to adjust the coefficients of the reactants and products so that each atom type is equal on both sides.

For instance, if you start with one carbon in methane, you'll also need one carbon in carbon disulfide. Each change affects the other elements so it often takes a few tries to balance it correctly. In this case, we balanced the equation as \(\mathrm{CH}_{4} + 4 ext {S}\rightarrow CS_{2} + 2 ext {H}_{2} ext{S}\).

Remember, practice makes perfect! The more you practice writing and balancing equations, the better you'll become.

Limiting Reagent

The concept of a limiting reagent is central to understanding how chemical reactions proceed. In a chemical reaction, the limiting reagent is the reactant that is completely consumed first, stopping the reaction and thus determining the maximum amount of product formed.

In this problem, we started with 120 g each of methane and sulfur. The key is to convert these masses into moles, as reactions occur in terms of moles, not mass. We calculated the moles of methane (\(\approx 7.50 \text{mol}\)) and sulfur (\(\approx 3.74\text{mol}\)) using their molar masses.

Next, we compared the actual mole ratio from the given masses to the ratio required by the balanced equation. The equation tells us we need four times as much sulfur as methane (\(1:4\)). But we only have roughly twice as much methane as sulfur (\(2:1\)). This mismatch shows that methane is our limiting reagent, as it will be used up before all the sulfur can be reacted.

Identifying the limiting reagent is critical as it directly impacts the amount of product that can form.

Theoretical Yield

Once you know your limiting reagent, you can predict the theoretical yield of the product, which is the maximum amount of product that could be formed from the reactants.

In our reaction, the balanced equation shows that each mole of methane (\(\mathrm{CH}_{4}\)) produces one mole of carbon disulfide (\(\mathrm{CS}_{2}\)). We have 7.50 moles of methane according to our calculations, meaning potentially 7.50 moles of carbon disulfide could be formed, assuming complete reaction.

To find the mass of carbon disulfide, we convert the moles back to grams using the molar mass of carbon disulfide, which is 76.15 g/mol:

• 7.50 moles \(\times \) 76.15 g/mol = 571.13 grams

Thus, the theoretical yield of carbon disulfide is approximately 571.13 grams,
assuming no loss in the process. Understanding theoretical yield helps in evaluating the efficiency of a reaction and anticipating how much product can be practically obtained.