Question

A favorite demonstration among chemistry instructors, to show that the properties of a compound differ from those of its constituent clements, involves iron filings and powdered sulfur. If the instructor takes samples of iron and sulfur and just mixes them togcther, the two elements can be separated from one another with a magnet (iron is attracted to a magnet, sulfur is not ). If the instructor then combines and heats the mixture of iron and sulfur, a reaction takes place and the elements combine to form iron(II) sulfide (which is not attracted by a magnet). $$ \mathrm{Fe}(s)+\mathrm{S}(s) \rightarrow \operatorname{FeS}(s) $$ Suppose \(5.25 \mathrm{~g}\) of iron filings is combined with \(12.7 \mathrm{~g}\) of sulfur. What is the theoretical yield of iron(ll) sulfidc?

Short answer

The theoretical yield of iron(II) sulfide when 5.25 g of iron filings is combined with 12.7 g of sulfur is 8.27 grams.

Step-by-step solution

Convert grams to moles

First, we need to convert the amounts of iron and sulfur given in grams to moles. To do this, we will use the molar masses of iron (Fe) and sulfur (S). Molar mass of Fe = 55.85 g/mol Molar mass of S = 32.07 g/mol Moles of Fe = (5.25 g) / (55.85 g/mol) = 0.0941 mol Moles of S = (12.7 g) / (32.07 g/mol) = 0.396 mol

Determine the limiting reactant

Next, we need to determine which reactant is the limiting reactant. To do this, we will compare the mole ratio of the reactants given (iron and sulfur) to the stoichiometry indicated in the balanced chemical equation. From the balanced chemical equation, the stoichiometric ratio of iron to sulfur is 1:1. We will calculate the ratio of moles of iron to moles of sulfur: Ratio of Fe to S = Moles of Fe / Moles of S = 0.0941 mol / 0.396 mol = 0.2376 Since the ratio of Fe to S (0.2376) is less than the stoichiometric ratio (1), Fe is the limiting reactant.

Calculate the theoretical yield of iron(II) sulfide

Now we can use stoichiometry to find the theoretical yield of iron(II) sulfide. For every mole of Fe reacting, 1 mole of FeS is produced. So, we will multiply the moles of the limiting reactant (Fe) by the stoichiometric conversion factor (1 mol FeS / 1 mol Fe) to find the moles of FeS produced. Moles of FeS = (0.0941 mol Fe) x (1 mol FeS / 1 mol Fe) = 0.0941 mol FeS Now, we just need to convert moles of FeS to grams using the molar mass of FeS: Molar mass of FeS = (Molar mass of Fe) + (Molar mass of S) = 55.85 g/mol + 32.07 g/mol = 87.92 g/mol Theoretical yield of FeS = (0.0941 mol FeS) x (87.92 g/mol) = 8.27 g The theoretical yield of iron(II) sulfide is 8.27 grams.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, meaning that matter cannot be created or destroyed in a closed system. Understanding stoichiometry helps students determine how much of each substance is required or produced in a reaction.
To solve stoichiometry problems, we mostly rely on balanced chemical equations, which provide the mole ratios of the reactants and products. These ratios are essential as they guide us in determining how much of one reactant is needed to react with another, and how much product will be formed.
In the exercise given, the balanced chemical equation for the reaction between iron (Fe) and sulfur (S) to form iron(II) sulfide (FeS) is:\[ \mathrm{Fe}(s) + \mathrm{S}(s) \rightarrow \mathrm{FeS}(s) \]This equation tells us that one mole of iron reacts with one mole of sulfur to produce one mole of iron sulfide. Therefore, the stoichiometric ratio of iron to sulfur to iron(II) sulfide is 1:1:1. This proportion allows us to determine the limiting reactant, which is the reactant that runs out first in a reaction, thus limiting the amount of product that can be formed.

Molar Mass

Molar mass is a key concept in chemistry that refers to the mass of one mole of a given substance, expressed in grams per mole (g/mol). Each element has a unique molar mass that is equal to its atomic weight listed on the periodic table.
In your exercise, the conversion from grams to moles is made possible by using the molar masses of iron (Fe) and sulfur (S). Specifically, the molar mass of iron is 55.85 g/mol, and that of sulfur is 32.07 g/mol.

• To find the moles of iron, you divide the mass of iron by its molar mass: \( \frac{5.25 \, \text{g}}{55.85 \, \text{g/mol}} \). This gives approximately 0.0941 mol of iron.
• Similarly, for sulfur, you do \( \frac{12.7 \, \text{g}}{32.07 \, \text{g/mol}} \), resulting in 0.396 mol of sulfur.

These calculations are crucial for determining which reactant is in excess and which is the limiting reactant. Understanding molar mass enables students to transition from the macroscopic scale of grams to the microscopic scale of moles, facilitating accurate predictions of reaction outcomes.

Theoretical Yield

Theoretical yield is the maximum amount of product that can be generated from a given amount of reactants in a chemical reaction, under ideal conditions and with complete conversion. Calculating the theoretical yield involves using the moles of the limiting reactant along with stoichiometric ratios from the balanced equation.
In the provided exercise, since iron (Fe) was determined to be the limiting reactant, it sets the maximum amount of iron(II) sulfide (FeS) that can be produced. For every mole of Fe, one mole of FeS is formed, as indicated by the stoichiometry of the reaction. Hence, the moles of Fe directly equate to moles of FeS produced.To convert moles of FeS to grams, we use its molar mass, which is calculated by adding the molar masses of iron and sulfur:\[ 87.92 \, \text{g/mol} \quad \text{(55.85 g/mol of Fe + 32.07 g/mol of S)} \]By multiplying the moles of FeS (0.0941 mol) by the molar mass of FeS (87.92 g/mol), we find the theoretical yield of FeS to be 8.27 grams. This value represents the upper limit of product formation, assuming all of the limiting reactant is used up.