Question

Calculate the theoretical yield of product (in moles) for each initial amount of reactants. $$ 3 \mathrm{Mn}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Mn}_{3} \mathrm{O}_{4}(s) $$ a. \(3 \mathrm{~mol} \mathrm{Mn}, 3 \mathrm{~mol} \mathrm{O}_{2}\) b. \(4 \mathrm{~mol} \mathrm{Mn}, 7 \mathrm{~mol} \mathrm{O}_{2}\) c. \(27.5 \mathrm{~mol} \mathrm{Mn}, 43.8 \mathrm{~mol} \mathrm{O}_{2}\)

Short answer

Theoretical yields: (a) 1.5 moles of Mn3O4, (b) 1.33 moles of Mn3O4, (c) 9.16 moles of Mn3O4.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical reaction is already given: 3 Mn(s) + 2 O2(g) → Mn3O4(s).

Calculate theoretical yield for 3 moles of Mn and 3 moles of O2

From the balanced reaction, 3 moles of Mn react with 2 moles of O2 to form 1 mole of Mn3O4.For 3 moles of Mn, the limiting reactant is O2 as it will be completely consumed first to form (3 moles O2) x (1 mole Mn3O4 / 2 moles O2) = 1.5 moles of Mn3O4.

Calculate theoretical yield for 4 moles of Mn and 7 moles of O2

Using the stoichiometry of the reaction, 4 moles of Mn would require 2.66 moles of O2 to form Mn3O4.Since there is an excess of O2 (7 moles), the theoretical yield is limited by Mn and is (4 moles Mn) x (1 mole Mn3O4 / 3 moles Mn) = 1.33 moles of Mn3O4.

Calculate theoretical yield for 27.5 moles of Mn and 43.8 moles of O2

By stoichiometry, 27.5 moles of Mn would require 18.33 moles of O2. Given 43.8 moles of O2, Mn is the limiting reactant.The theoretical yield is (27.5 moles Mn) x (1 mole Mn3O4 / 3 moles Mn) = 9.16 moles of Mn3O4.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. Thus, the amount of reactants consumed will be equal to the amount of products formed, following a fixed ratio dictated by the balanced chemical equation.

Understanding stoichiometry is crucial for determining how much of each reactant is needed and how much product will be formed in a chemical reaction. To perform a stoichiometry calculation, one must begin with a balanced chemical equation and use it as a roadmap to relate the quantities of reactants and products using molar ratios. These ratios are derived from the coefficients of the balanced equation and reflect the proportions in which chemicals react.

For educational purposes, let's consider the reaction given in the exercise:
\(3 \text{Mn}(s) + 2 \text{O}_{2}(g) \rightarrow \text{Mn}_{3}\text{O}_{4}(s)\).
This indicates that three moles of manganese (Mn) react with two moles of oxygen gas (O2) to produce one mole of manganese(II,III) oxide (Mn3O4). Stoichiometric calculations often begin by identifying the molar amounts of reactants, converting them into moles, if necessary, and then using the molar ratios from the balanced equation to calculate the amount of product.

Limiting Reactant

The concept of the limiting reactant is an essential aspect of stoichiometry and theoretical yield calculations. In a chemical reaction, the limiting reactant is the substance that is completely consumed first, therefore determining the maximum amount of product that can be formed. Once the limiting reactant is used up, the reaction stops, and no further product is produced, even if other reactants are still present in excess.

To identify the limiting reactant, one must compare the mole ratios of the reactants used with the mole ratios in the balanced chemical equation. The reactant that has a lesser amount, in terms of the stoichiometric ratio, is the limiting reactant. In the exercise provided, the balanced equation is used to evaluate which reactant will limit the formation of Mn3O4 for different initial mole quantities of Mn and O2.

For instance, in scenario (a) with 3 moles of Mn and 3 moles of O2, oxygen is the limiting reactant. This is concluded based on the reaction stoichiometry where 2 moles of O2 yield 1 mole of Mn3O4. Even if there are equal molar amounts of Mn and O2, the ratio of O2 to Mn in the balanced equation shows that O2 will be consumed faster. Therefore, one can calculate the theoretical yield of Mn3O4 based on the available moles of O2.

Chemical Reaction Balancing

Balancing chemical equations is a fundamental skill in chemistry that entails adjusting the coefficients of reactants and products to ensure that the number of atoms for each element is conserved across both sides of the equation. This reflects the principle of the conservation of mass and is vital for stoichiometry calculations.

To balance a chemical equation, one typically follows several steps:

Identify Unbalanced Elements

First, list all the elements involved and count the number of atoms for each element on the reactant and product sides.

Balance Elements One by One

Begin with elements that appear in a single reactant and product. Adjust coefficients to achieve the same number of atoms on both sides.

Balance Polyatomic Ions as a Whole

If the same polyatomic ion appears on both sides, treat it as a single unit.

Consider the Conservation of Charge

If the reaction involves ions, ensure that the overall charge is balanced as well.

Once an equation is balanced, like the provided equation \(3 \text{Mn}(s) + 2 \text{O}_{2}(g) \rightarrow \text{Mn}_{3}\text{O}_{4}(s)\), it serves as the foundation for all subsequent stoichiometric computations. A balanced equation applies universally, meaning it accurately represents the relative amounts of reactants and products involved, regardless of how much of each substance is present initially. This is why ensuring an equation is correctly balanced before performing any further calculations is critical.