Problem 14 - Chemistry and Chemical Reactivity

Question

Disulfur dichloride, \(S_{2} C l_{2},\) is used to vulcanize rubber. It can be made by treating molten sulfur with gaseous chlorine: $$\mathrm{S}_{8}(\ell)+4 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{s}_{2} \mathrm{Cl}_{2}(\ell)$$ Starting with a mixture of \(32.0 \mathrm{g}\) of sulfur and \(71.0 \mathrm{g}\) of \(\mathrm{Cl}_{2}\). (a) Which is the limiting reactant? (b) What is the theoretical yield of \(S_{2} C l_{2} ?\) (c) What mass of the excess reactant remains when the reaction is completed?

Step-by-Step Solution

Answer

(a) \(S_8\) is the limiting reactant. (b) The theoretical yield of \(S_2Cl_2\) is \(67.52\, g\). (c) \(35.50\, g\) of \(Cl_2\) remains.

Step 1: Determine Moles of Each Reactant

First, we need to calculate the moles of sulfur and chlorine gas. The molar mass of sulfur, \(S_{8}\), is approximately \(256.52 \, g/mol\) (8 times the atomic mass of sulfur which is about \(32.07 \, g/mol\)). The molar mass of \(Cl_{2}\) is approximately \(70.90 \, g/mol\) (2 times the atomic mass of chlorine which is about \(35.45 \, g/mol\)).\[\text{Moles of } S_8 = \frac{32.0 \, g}{256.52 \, g/mol} = 0.125 \, mol\]\[\text{Moles of } Cl_2 = \frac{71.0 \, g}{70.90 \, g/mol} = 1.001 \, mol\]

Step 2: Identify Limiting Reactant

Using the stoichiometry of the balanced equation \(\mathrm{S}_{8} + 4 \mathrm{Cl}_{2} \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}\), we see that 1 mole of \(S_8\) reacts with 4 moles of \(Cl_2\).To react all \(0.125 \, mol\) of \(S_8\), we need:\[4 \times 0.125 = 0.500 \, mol\] of \(Cl_2\).Since we have \(1.001 \, mol\) of \(Cl_2\), chlorine is in excess, and sulfur \(S_8\) is the limiting reactant.

Step 3: Calculate Theoretical Yield of \(S_2Cl_2\)

The balanced equation shows that 1 mole of \(S_8\) produces 4 moles of \(S_2Cl_2\). From 0.125 moles of \(S_8\), we can produce:\[0.125 \, mol \times 4 = 0.500 \, mol\] of \(S_2Cl_2\).The molar mass of \(S_2Cl_2\) is approximately \(135.04 \, g/mol\) (2 times sulfur and 2 times chlorine).Thus, the theoretical yield of \(S_2Cl_2\) is:\[0.500 \, mol \times 135.04 \, g/mol = 67.52 \, g\]

Step 4: Calculate Mass of Excess Reactant Remaining

Calculate the moles of \(Cl_2\) that actually react with \(S_8\). We already know that \(0.125 \, mol\) of \(S_8\) requires \(0.500 \, mol\) of \(Cl_2\).Therefore, the moles of \(Cl_2\) that remain:\[1.001 \, mol - 0.500 \, mol = 0.501 \, mol\]The mass of excess \(Cl_2\) is given by:\[0.501 \, mol \times 70.90 \, g/mol = 35.50 \, g\]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand how to measure and relate the quantities of reactants and products in a chemical reaction. It utilizes the balanced chemical equation to determine these relationships. By looking at the specific ratios defined by the coefficients of the equation, you can predict the amount of products formed from given reactants. For our reaction: \[\mathrm{S}_{8} + 4 \mathrm{Cl}_{2} \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}\]- One mole of \(\mathrm{S}_{8}\) will react with four moles of \(\mathrm{Cl}_{2}\) to produce four moles of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\). - This ratio helps determine the limiting reactant because it tells us exactly how much of each reactant is needed to ensure complete consumption. - Stoichiometry acts like a recipe guide, ensuring we have just the right amount of ingredients to "bake" our chemical "cake" without any wastage.

Theoretical Yield

Theoretical yield represents the maximum amount of product you could obtain from a reaction, assuming complete conversion of the limiting reactant with no losses. This is based on the stoichiometric relationships in the chemical equation. In our exercise:- Sulfur \(\mathrm{S}_{8}\) is the limiting reactant, meaning it will dictate the maximum amount of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) produced. - To find the theoretical yield, calculate how many moles of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) can be produced from the \(0.125\, \text{mol}\) of \(\mathrm{S}_{8}\). According to the equation, each mole of \(\mathrm{S}_{8}\) creates four moles of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\).In this exercise, we calculated that \(0.125 \text{ mol}\) of \(\mathrm{S}_{8}\) yields \(0.500 \text{ mol}\) of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\). With the molar mass of \(135.04 \text{ g/mol}\), this results in a theoretical yield of \(67.52 \text{ g}\) of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\).

Molar Mass

Molar mass is the weight in grams of one mole of a substance. It connects the macroscopic and molecular scales, allowing us to convert between the mass of a substance and the number of moles. To find molar mass:- Add up the atomic weights of all atoms in the formula based on the periodic table.In the exercise:- Sulfur \(\mathrm{S}_{8}\) has a molar mass of \(256.52 \text{ g/mol}\), calculated by multiplying the atomic mass of sulfur \(32.07 \text{ g/mol}\) by 8.- For chlorine gas \(\mathrm{Cl}_{2}\), the molar mass is \(70.90 \text{ g/mol}\), as chlorine has an atomic mass of \(35.45 \text{ g/mol}\) and there are two atoms in \(\mathrm{Cl}_{2}\).Correctly calculating molar mass is crucial for converting grams to moles, which then feeds into stoichiometry calculations.

Excess Reactant

The excess reactant is the substance that remains after a chemical reaction has completed. It does not fully react because there is more of it available than needed to consume the limiting reactant. Understanding how much excess remains can be important to determine economic efficiency or to plan for recycling or disposal.In our scenario:- Chlorine \(\mathrm{Cl}_{2}\) is in excess. Starting with \(1.001 \text{ mol}\) of \(\mathrm{Cl}_{2}\), only \(0.500 \text{ mol}\) reacts with the sulfur.- This means \(0.501 \text{ mol}\) of \(\mathrm{Cl}_{2}\) does not react and remains unconsumed.To find the mass of the excess \(\mathrm{Cl}_{2}\), multiply the moles of excess by its molar mass, resulting in \(35.50 \text{ g}\) that remains after the reaction.