Question

Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4} .\) They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes.$$\begin{array}{c}\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\end{array}$$

Short answer

The balanced equations for the given reactions are: 1. \( \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+2 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) 2. \( \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{CO}(g) \rightarrow 3 \mathrm{Fe}(s)+4 \mathrm{CO}_{2}(g)\)

Step-by-step solution

Count atoms on both sides

First, let's count how many atoms of each element are present on both sides of the equation. Unbalanced equation: \[\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g)\] Atoms on the left side: - 3 Fe - 4 O - 2 H Atoms on the right side: - 1 Fe - 2 O - 2 H

Balance Fe atoms

On the right side, there's only one Fe atom, while there are three on the left side. So, let's put a 3 in front of Fe on the right side to balance the Fe atoms. \[\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow 3 \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g)\]

Balance O atoms

Now, there are 2 O atoms on the right side and 4 O atoms on the left side. To balance the O atoms, let's put a 2 in front of H2O on the right side. \[\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow 3 \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\]

Balance H atoms

Finally, we need to balance the H atoms. There are currently 4 H atoms on the right side and 2 H atoms on the left side. So, let's put a 2 in front of H2 on the left side. \[ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+2 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\] Now, the first equation is balanced. For the equation with carbon monoxide:

Count atoms on both sides

First, let's count how many atoms of each element are present on both sides of the equation. Unbalanced equation: \[\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\] Atoms on the left side: - 3 Fe - 4 O - 1 C Atoms on the right side: - 1 Fe - 2 O - 1 C

Balance Fe atoms

On the right side, there's only one Fe atom, while there are three on the left side. So, let's put a 3 in front of Fe on the right side to balance the Fe atoms. \[\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \rightarrow 3 \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]

Balance O atoms

Now, there are 3 O atoms on the left side (Fe3O4 has 4 O atoms) and 2 O atoms on the right side. To balance the O atoms, we need to put a 4 in front of CO on the left side. \[ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{CO}(g) \rightarrow 3 \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\]

Balance C atoms

Finally, we need to balance the C atoms. There are currently 4 C atoms on the left and only 1 C atom on the right side. So, let's put a 4 in front of CO2 on the right side. \[ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{CO}(g) \rightarrow 3 \mathrm{Fe}(s)+4 \mathrm{CO}_{2}(g)\] Now, the second equation is balanced. So, the balanced equations are: \[ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+2 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\] \[ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{CO}(g) \rightarrow 3 \mathrm{Fe}(s)+4 \mathrm{CO}_{2}(g)\]

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative aspects of chemical reactions. It allows us to calculate the amounts of reactants and products involved in a chemical reaction based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means that for a chemical equation to accurately represent a reaction, the number of atoms of each element must be the same on both sides of the equation.

Let's consider the example given. To balance the equation involving iron oxide and hydrogen, we had to ensure that the number of iron (Fe), oxygen (O), and hydrogen (H) atoms are equal on both sides. Here, stoichiometry plays a crucial role as it guides us through the steps needed to balance the equation: finding the correct coefficients (the numbers placed before compounds) to ensure the same number of atoms of each element is present on both sides of the reaction.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to form new substances. The substances that undergo the chemical change are called reactants, and the new substances formed are called products. Chemical reactions are represented by chemical equations, which have reactants on the left side, products on the right side, and an arrow indicating the direction of the reaction.

In the given exercise, we examine two types of reactions involving iron oxide: its reaction with hydrogen gas and with carbon monoxide. Each iron oxide molecule reacts with hydrogen or carbon monoxide molecules to produce iron and water or carbon dioxide, respectively. The essence of learning about chemical reactions lies in understanding how to use symbols and formulas to represent them and how to predict the products of a reaction.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are a type of chemical reaction that involves the transfer of electrons between substances. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. In a redox reaction, one substance is oxidized, and another is reduced.

In our exercise, the reaction between iron oxide and carbon monoxide can be viewed as a redox process. Iron oxide is reduced to iron metal, meaning it gains electrons, and carbon monoxide is oxidized to carbon dioxide, meaning it loses electrons. By examining such reactions, students gain insight into the electron transfer processes that are fundamental to chemical reactions, battery operation, and corrosion prevention, among other applications.