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: $$\begin{array}{c}\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\end{array}$$

Short answer

The balanced equations are: \[ \renewcommand*{\arraystretch}{1.4}\begin{array}{c} \mathrm{Fe_{3}O_{4}}(s) + 4\mathrm{H_{2}}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H_{2}O}(g) \\ \mathrm{Fe_{3}O_{4}}(s) + 4\mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{CO_{2}}(g) \end{array} \]

Step-by-step solution

We are given the equation, \[ \mathrm{Fe_{3}O_{4}}(s) + \mathrm{H_{2}}(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{H_{2}O}(g).\] Count the atoms on both sides for each element: Fe, O, and H. #Step 2: Balancing Fe atoms in the first equation#

There are 3 Fe atoms on the left side and 1 Fe atom in the right side. To balance this, place a 3 in front of the \(\mathrm{Fe}\): \[ \mathrm{Fe_{3}O_{4}}(s) + \mathrm{H_{2}}(g) \longrightarrow 3\mathrm{Fe}(s) + \mathrm{H_{2}O}(g). \] #Step 3: Balancing O and H atoms in the first equation#

Now, there are 4 O atoms on the left side and 1 O atom in the right side. To balance this, place a 4 in front of \(\mathrm{H_{2}O}\), and then balance H atoms as well: \[ \mathrm{Fe_{3}O_{4}}(s) + \cancelto{8}{4}\mathrm{H_{2}}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H_{2}O}(g). \] #Step 4: Analyzing the second equation#

We are given the equation, \[ \mathrm{Fe_{3}O_{4}}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{CO_{2}}(g).\] Count the atoms on both sides for each element: Fe, O, and C. #Step 5: Balancing Fe atoms in the second equation#

There are 3 Fe atoms on the left side and 1 Fe atom in the right side. To balance this, place a 3 in front of the \(\mathrm{Fe}\): \[ \mathrm{Fe_{3}O_{4}}(s) + \mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + \mathrm{CO_{2}}(g). \] #Step 6: Balancing O and C atoms in the second equation#

Now, there are 4 O atoms on the left side and 2 O atoms in the right side. To balance this, place a 4 in front of \(\mathrm{CO}\), and then balance C atoms as well: \[ \cancel{4}\mathrm{Fe_{3}O_{4}}(s) + 4\mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{CO_{2}}(g). \] The balanced equations are: \[ \renewcommand*{\arraystretch}{1.4}\begin{array}{c} \mathrm{Fe_{3}O_{4}}(s) + 4\mathrm{H_{2}}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H_{2}O}(g) \\ \mathrm{Fe_{3}O_{4}}(s) + 4\mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{CO_{2}}(g) \end{array} \]

Key concepts

Iron Oxide Reduction

When we talk about iron oxide reduction, we're describing a process that is central to the extraction of iron from its ores. This process involves removing the oxygen atoms that are bound to iron atoms in the ore. In the exercise given, both carbon monoxide (CO) and hydrogen gas (H₂) are used as reducing agents to transform iron oxide (Fe₃O₄) into elemental iron (Fe).

This reaction is critical in metallurgy and materials chemistry because it's a fundamental step in producing iron, which can be further refined and used to manufacture steel and various other alloys. The balanced chemical equations provide a clear stoichiometric representation of the redox reaction that occurs when the iron oxide is reduced.

Stoichiometry

The term stoichiometry refers to the calculation of the quantities of reactants and products in chemical reactions. In essence, it's the 'recipe' for a chemical reaction—describing the relative amounts of each substance involved.

In illustrating the stoichiometry of the reduction of iron oxide, it is essential to balance chemical equations to obey the law of conservation of mass. This means that the number of atoms of each element must be the same on both sides of the equation. In the given examples, coefficients are placed in front of the chemical formulas to ensure that the iron (Fe), oxygen (O), and reducing agent atoms balance in the total reaction.

Redox Reactions

A redox reaction is a type of chemical reaction that involves a transfer of electrons between two species. It is short for 'reduction-oxidation reaction.' The substance that loses electrons is oxidized, and the one that gains electrons is reduced.

In the context of the reduction of iron oxide, the iron oxide is reduced as it gains electrons from the reducing agent, which is itself oxidized. This electron transfer is what drives the formation of elemental iron from its oxide. Redox reactions are fundamental in chemistry and are harnessed in various applications, including energy storage (such as batteries) and in industrial processes like the one described in our exercise.

Mole Concept

The mole concept is a bridge between the macroscopic world that we can observe directly and the microscopic world of atoms and molecules. A mole is a unit in chemistry that represents a specific amount of substance—approximately 6.022 x 10²³ particles of that substance.

The solution presented uses the mole concept to ensure that the number of molecules reacting (and produced) in the equation is accounted for correctly. For instance, 4 moles of H₂ react with 1 mole of Fe₃O₄ to produce 3 moles of Fe and 4 moles of H₂O. Understanding the mole concept is essential in converting between atom counts to more practical, measurable quantities in grams and liters, enabling chemists to scale reactions up from the level of atoms to the laboratory or industrial scale.