Problem 13 - Chemistry and Chemical Reactivity

Question

The compound \(\mathrm{SF}_{6}\) is made by burning sulfur in an atmosphere of fluorine. The balanced equation is $$\mathrm{S}_{8}(\mathrm{s})+24 \mathrm{F}_{2}(\mathrm{g}) \rightarrow 8 \mathrm{SF}_{6}(\mathrm{g})$$ Starting with a mixture of 1.6 mol of sulfur, \(S_{8,}\) and 35 mol of \(\mathrm{F}_{2}\), (a) Which is the limiting reagent? (b) What amount of \(\mathrm{SF}_{6}\) is produced?

Step-by-Step Solution

Answer

F2 is the limiting reagent; 11.67 moles of SF6 is produced.

Step 1: Determine Molar Ratios from the Balanced Equation

The balanced chemical equation is \( \mathrm{S}_{8} + 24 \mathrm{F}_{2} \rightarrow 8 \mathrm{SF}_{6} \). From this equation, 1 mole of \( \mathrm{S}_{8} \) reacts with 24 moles of \( \mathrm{F}_{2} \) to produce 8 moles of \( \mathrm{SF}_{6} \).

Step 2: Compare Given Moles to Required Moles for Reaction

We have 1.6 mol of \( \mathrm{S}_{8} \) and 35 mol of \( \mathrm{F}_{2} \). Using the balanced equation ratios, determine the number of moles of \( \mathrm{F}_{2} \) required to react with 1.6 mol of \( \mathrm{S}_{8} \): \( 1.6 \text{ moles of } \mathrm{S}_{8} \times \frac{24 \text{ moles of } \mathrm{F}_{2}}{1 \text{ mole of } \mathrm{S}_{8}} = 38.4 \text{ moles of } \mathrm{F}_{2} \).

Step 3: Identify Limiting Reagent

Compare the required moles of \( \mathrm{F}_{2} \) to the moles available. We require 38.4 moles of \( \mathrm{F}_{2} \), but only have 35 moles available. Thus, \( \mathrm{F}_{2} \) is the limiting reagent.

Step 4: Calculate Yield of \( \mathrm{SF}_{6} \)

Since \( \mathrm{F}_{2} \) is the limiting reagent, calculate how much \( \mathrm{SF}_{6} \) can be produced. Using 35 moles of \( \mathrm{F}_{2} \), \( 35 \text{ moles of } \mathrm{F}_{2} \times \frac{8 \text{ moles of } \mathrm{SF}_{6}}{24 \text{ moles of } \mathrm{F}_{2}} = 11.67 \text{ moles of } \mathrm{SF}_{6} \), which is the amount of \( \mathrm{SF}_{6} \) produced.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. It uses the balanced chemical equation to ensure reactions occur in the correct proportions, allowing us to determine how much of each substance is involved. This concept helps us understand and calculate the necessary amounts of reactants needed to produce a desired amount of product. By using stoichiometry, we are able to determine the limiting reagent, which is the reactant that is completely consumed first and limits the amount of product formed. In our example with \(\mathrm{S}_{8}\) and \(\mathrm{F}_{2}\), stoichiometry guides us in calculating that \(\mathrm{F}_{2}\) is the limiting reagent because it falls short compared to the stoichiometric requirement specified in the balanced equation.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into new substances called products. During this transformation, bonds between atoms are broken and new bonds are formed, resulting in different chemical entities. A balanced chemical equation represents a chemical reaction, indicating the reactants and products involved along with their relative amounts. Balancing chemical equations is crucial because it reflects the conservation of mass, energy, and charge. In the equation given for the formation of sulfur hexafluoride (\(\mathrm{SF}_6\)), \(\mathrm{S}_{8}\) reacts with \(\mathrm{F}_{2}\) to form \(\mathrm{SF}_{6}\). The challenge often lies in discovering which reactant limits the progression of the reaction and dictates the quantity of product formed.

Molar Ratios

Molar ratios are the proportions of reactants and products involved in a chemical reaction as indicated by a balanced equation. They are essential for solving stoichiometry problems. A balanced equation shows the number of moles of each reactant needed to produce a given number of moles of product. For instance, the balanced reaction \(\mathrm{S}_{8} + 24\ \mathrm{F}_{2} \rightarrow 8\ \mathrm{SF}_{6}\) shows that 1 mole of \(\mathrm{S}_{8}\) requires 24 moles of \(\mathrm{F}_{2}\) to produce 8 moles of \(\mathrm{SF}_6\). This ratio is the key to identifying the limiting reagent and calculating the amount of product formed. By comparing the available moles of each reactant to these molar ratios, we can determine the maximum yield of a product. In our example, the available 35 moles of \(\mathrm{F}_{2}\) can be compared to the required 38.4 moles, revealing that \(\mathrm{F}_{2}\) is insufficient and thus the limiting reagent.