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.

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: \(CH_4 + 4S \rightarrow CS_2 + 2H_2S\)

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.

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.