Question

Apply In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(lill) sulfide. $$2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{S}(\mathrm{s}) \rightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})$$ What is the theoretical yield, in grams, of iron(III) sulfide?

Short answer

The theoretical yield of iron(III) sulfide in this experiment is approximately 155.90 grams.

Step-by-step solution

Calculate moles of iron.

We will convert grams of iron to moles using the molar mass of iron: 1 mol of Fe = 55.85 g Moles of Fe = (83.77 g Fe) / (55.85 g Fe/mol) ≈ 1.50 mol Fe

Compute moles of iron(III) sulfide.

Using stoichiometry and the mole ratio from the provided balanced chemical equation, we will determine the moles of iron(III) sulfide produced: Fe to Fe₂S₃ ratio = 2 : 1 Moles of Fe₂S₃ = (1.50 mol Fe) * (1 mol Fe₂S₃ / 2 mol Fe) = 0.750 mol Fe₂S₃

Convert moles of iron(III) sulfide to grams.

We'll now convert the moles of iron(III) sulfide to grams, using the molar mass of iron(III) sulfide: 1 mol Fe₂S₃ = (2 * 55.85 g Fe) + (3 * 32.07 g S) = 207.87 g Theoretical yield of Fe₂S₃ = (0.750 mol Fe₂S₃) * (207.87 g Fe₂S₃/mol) ≈ 155.90 g Fe₂S₃ The theoretical yield of iron(III) sulfide is approximately 155.90 grams.

Key concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that involves the calculation of reactants and products in chemical reactions. Simply put, it is the math behind how substances combine and change in a reaction. This concept helps scientists determine the right amounts of each substance needed or produced in a reaction. In this exercise, stoichiometry was used to calculate the theoretical yield of iron(III) sulfide, which is the amount of product that should be created from a given amount of reactant under ideal conditions.
Think of stoichiometry like following a recipe: If you follow the measurements exactly, you'll get the expected result. Similarly, in chemistry, using the correct proportions of reactants (based on mole ratios) ensures accurate predictions of product amounts.

Moles

Chemists use the mole as a unit to quantify the amount of a substance. One mole contains exactly Avogadro's number, which is approximately \(6.022 \times 10^{23}\) of some chemical unit, such as atoms or molecules. This large number is handy because substances at the molecular level are incredibly small.
In the context of this exercise, moles helped in converting the mass of iron into a measurable amount that could be applied in stoichiometric calculations. The conversion was simple: using the molar mass of iron, which is the mass of one mole of iron atoms. In doing so, it becomes easier to determine how much iron is reacting and, subsequently, how much product could be formed.

Balanced Chemical Equation

A balanced chemical equation is essential to accurately perform stoichiometric calculations. It ensures that matter is conserved during the reaction, meaning the same number of each type of atom appears on both sides of the reaction equation.
For the reaction involving iron and sulfur to form iron(III) sulfide, the equation is balanced as follows:

• \[2 \mathrm{Fe} + 3 \mathrm{S} \rightarrow \mathrm{Fe}_2 \mathrm{S}_3\]

This equation reflects the stoichiometric relationship between reactants and products, showing the ratio in which substances react and are produced. For instance, two moles of iron react with three moles of sulfur to produce one mole of iron(III) sulfide, providing a roadmap for mathematical conversions based on mole ratios.

Molar Mass

Molar mass is critical in converting between grams and moles in chemical equations. It is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). Molar mass allows us to translate between the mass we measure and the moles used in calculations.
In this exercise, the molar mass of both iron and iron(III) sulfide was pivotal for conversions. The molar mass of iron, \(55.85 \text{ g/mol}\), was used to determine how many moles of iron were available to react. Then, the molar mass of iron(III) sulfide, \(207.87 \text{ g/mol}\), helped compute the mass of the theoretical yield from its moles.
Understanding molar mass helps you see how much of a compound will form from a given mass of a reactant, bridging the reactants' measurements and the products' quantities.