Problem 14 - Chemistry and Chemical Reactivity

Question

Disulfur dichloride, \(\mathrm{S}_{2} \mathrm{Cl}_{2}\), is used to vulcanize rubber. It can be made by treating molten sulfur with gaseous chlorine: $$ \mathrm{S}_{\mathrm{s}}(\ell)+4 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}(\ell) $$ 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 \(\mathrm{S}_{2} \mathrm{Cl}_{2} ?\) (c) What mass of the excess reactant remains when the reaction is completed?

Step-by-Step Solution

Answer

(a) \( \mathrm{Cl}_2 \) is the limiting reactant. (b) 33.76 g of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is the theoretical yield. (c) 24.05 g of sulfur remains.

Step 1 - Calculate Moles of Reactants

First, find the molar mass of sulfur, \( \mathrm{S} \), which is \( 32.07\, \mathrm{g/mol} \). Then, calculate the moles of sulfur: \( \frac{32.0\, \mathrm{g}}{32.07\, \mathrm{g/mol}} = 1.00\, \mathrm{mol} \).Next, find the molar mass of chlorine, \( \mathrm{Cl}_2 \), which is \( 70.9\, \mathrm{g/mol} \). Calculate the moles of chlorine: \( \frac{71.0\, \mathrm{g}}{70.9\, \mathrm{g/mol}} = 1.00\, \mathrm{mol} \).

Step 2 - Determine Limiting Reactant

The balanced equation is: \[ \mathrm{S}_{\mathrm{s}}(\ell) + 4 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{S}_{2} \mathrm{Cl}_{2}(\ell) \]According to the equation, \( 1 \) mole of sulfur reacts with \( 4 \) moles of \( \mathrm{Cl}_2 \). Here, \( 1.00 \) mole of sulfur would require \( 4.00 \) moles of \( \mathrm{Cl}_2 \), but we only have \( 1.00 \) mole of \( \mathrm{Cl}_2 \). Thus, \( \mathrm{Cl}_2 \) is the limiting reactant.

Step 3 - Calculate Theoretical Yield of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \)

Since \( \mathrm{Cl}_2 \) is the limiting reactant and \( \frac{1}{4} \) mole of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is produced per mole of \( \mathrm{Cl}_2 \), the theoretical yield is:\[ 1.00\, \text{mol} \times \frac{1}{4} = 0.25\, \text{mol of } \mathrm{S}_{2} \mathrm{Cl}_{2} \]The molar mass of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) is \( 135.04\, \mathrm{g/mol} \).Calculate the mass: \( 0.25\, \text{mol} \times 135.04\, \mathrm{g/mol} = 33.76\, \mathrm{g} \).

Step 4 - Calculate Mass of Excess Reactant Remaining

Initially, \( 1.00 \) mole of \( \mathrm{Cl}_2 \) reacts completely, requiring \( \frac{1.00}{4} = 0.25 \) moles of sulfur.The initial moles of sulfur are \( 1.00 \) mole, so the moles of sulfur left are \( 1.00 - 0.25 = 0.75 \) moles.Find the mass of the leftover sulfur: \( 0.75\, \text{mol} \times 32.07\, \mathrm{g/mol} = 24.05\, \mathrm{g} \).

Key Concepts

Mole Calculation

Mole calculation is a fundamental aspect of chemistry that helps in determining how much of each substance participates in a chemical reaction. When calculating moles, you need to know the molar mass of each reactant, which you can find on the periodic table. The molar mass provides the link between grams of a substance and the amount in moles. In our exercise example, we calculate the moles of sulfur and chlorine using their respective molar masses:

• For sulfur (\( \mathrm{S} \)), with a molar mass of \( 32.07\, \mathrm{g/mol} \)
• For chlorine (\( \mathrm{Cl}_2 \)), with a molar mass of \( 70.9\, \mathrm{g/mol} \)

By dividing the mass of each reactant by its molar mass, you calculate the moles involved in the reaction.To further understand and accurately perform such calculations, it's essential to be comfortable with converting between grams and moles. This step is crucial for determining how much of each chemical is needed and what remains after the reaction. Always check your calculations to ensure accuracy before solving further steps of a chemical reaction.

Theoretical Yield

The theoretical yield refers to the maximum amount of product that can be produced in a chemical reaction based on the stoichiometry and the limiting reactant. It is calculated by using the balanced chemical equation to determine the proportion of reactants involved. In our case, the limiting reactant is chlorine, \( \mathrm{Cl}_2 \), since we have less than what the stoichiometry requires.Once the limiting reactant is identified, the theoretical yield is computed by relating moles of the limiting reactant to moles of the desired product. According to the reaction equation, every 4 moles of \( \mathrm{Cl}_2 \) produces 1 mole of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).Given we have 1 mole of \( \mathrm{Cl}_2 \), it results in \( 0.25 \) moles of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).Finally, to find the mass of the theoretical yield, you multiply the moles of product by its molar mass, \( 135.04\, \mathrm{g/mol} \), which yields \( 33.76\, \mathrm{g} \) in this example. This process highlights how understanding stoichiometry and limiting reactants can predict the potential outcome of reactions before they are carried out in the lab.

Stoichiometry

Stoichiometry is the heart of balancing chemical reactions and allows chemists to predict how much of each substance will react, be produced, or remain. It is based on the balanced chemical equation, which uses relative quantities of reactants and products in moles.A stoichiometrically balanced equation ensures that the number of atoms for each element is equal on both sides of the equation. In this exercise, the equation shows that 1 mole of sulfur reacts with 4 moles of chlorine to produce 4 moles of \( \mathrm{S}_{2} \mathrm{Cl}_{2} \).When considering which reactant limits the reaction, stoichiometry is essential because it relies on these mole ratios. Identifying the limiting reactant—chlorine in this situation—helps determine how much \( \mathrm{S}_{2} \mathrm{Cl}_{2} \) can form and how much sulfur will be in excess after the reaction. Stoichiometry also helps calculate remaining quantities once the reaction is complete, allowing chemists to optimize processes, save resources, and minimize waste.