Question

Balance each of the following chemical equations by inspection: (a) \(\mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+\mathrm{H}_{2}(g)\) (b) \(\mathrm{FeS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{SO}_{2}(g)\)

Short answer

(a) 3 Fe(s) + 4 H2O(g) → Fe3O4(s) + 4 H2(g). (b) 4 FeS(s) + 7 O2(g) → 2 Fe2O3(s) + 4 SO2(g).

Step-by-step solution

Write the Unbalanced Equation (for a)

The unbalanced chemical equation is given as: \[ \mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) + \mathrm{H}_{2}(g) \]

Balance Iron Atoms (for a)

There are three iron atoms in \(\mathrm{Fe}_{3}\mathrm{O}_{4}\), so we need three \(\mathrm{Fe}\) atoms on the reactant side: \[ 3\ \mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) + \mathrm{H}_{2}(g) \]

Balance Hydrogen Atoms (for a)

There are two hydrogen atoms in \(\mathrm{H}_{2}\), so we need four water molecules to provide these hydrogen atoms: \[ 3\ \mathrm{Fe}(\mathrm{s}) + 4\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) + 4\ \mathrm{H}_{2}(g) \]

Balance Oxygen Atoms (for a)

The oxygen atoms are already balanced with the four water molecules providing four oxygen atoms for \(\mathrm{Fe}_{3}\mathrm{O}_{4}\). Thus the balanced equation for (a) is: \[ 3\ \mathrm{Fe}(\mathrm{s}) + 4\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) + 4\ \mathrm{H}_{2}(g) \]

Write the Unbalanced Equation (for b)

The unbalanced chemical equation is given as: \[ \mathrm{FeS}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{SO}_{2}(g) \]

Balance Iron Atoms (for b)

There are two iron atoms in \(\mathrm{Fe}_{2} \mathrm{O}_{3}\), so we need two \(\mathrm{FeS}\) units: \[ 2\ \mathrm{FeS}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) + \mathrm{SO}_{2}(g) \]

Balance Sulfur Atoms (for b)

Adding two units of \(\mathrm{SO}_{2}\) balances the sulfur: \[ 2\ \mathrm{FeS}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) + 2\ \mathrm{SO}_{2}(g) \]

Balance Oxygen Atoms (for b)

There are three oxygen atoms from \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and four from \(\mathrm{SO}_{2}\), totaling seven. Thus, 3.5 molecules of \(\mathrm{O}_{2}\) are needed: \[ 2\ \mathrm{FeS}(s) + 3.5\ \mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) + 2\ \mathrm{SO}_{2}(g) \]. Adjusting for integer coefficients, multiply by 2.

Final Balanced Equation (for b)

The balanced equation becomes: \[ 4 \ \mathrm{FeS}(s) + 7 \ \mathrm{O}_{2}(g) \rightarrow 2 \ \mathrm{Fe}_{2}\mathrm{O}_{3}(\mathrm{~s}) + 4 \ \mathrm{SO}_{2}(g) \].

Key concepts

Stoichiometry

Stoichiometry is fundamentally about the relationships between the quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the quantitative balancing of chemical equations.

To balance a chemical equation, you have to ensure that the number of each type of atom is conserved; this reflects the law of conservation of mass. For example, when balancing the equation \( 3 \ \mathrm{Fe}(\mathrm{s}) + 4 \ \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3}\mathrm{O}_{4}(\mathrm{~s}) + 4 \ \mathrm{H}_{2}(g) \), stoichiometry tells us that three iron atoms and four pairs of hydrogen are involved. The purpose here is to ensure not only that the amount of atoms is conserved but the proportion of the substances involved.

Key steps to remember in stoichiometry:

• Identify the number of each element's atoms present in the reactants and products.
• Adjust coefficients to ensure equal numbers of each type of atom on both sides of the equation.
• Use whole numbers when balancing equations, which might involve multiplying all coefficients to remove fractions.

Balancing chemical equations is a practical application of stoichiometry which ensures the equations are both quantifiable and predictable.

Chemical Reactions

At the core, chemical reactions involve the transformation of substances, the reactants, into new substances, the products. This change happens when chemical bonds are broken and new ones are formed, rearranging atoms into different configurations. Understanding the types of reactions is crucial for balancing equations and predicting reactions.

In our example: - In equation (a), iron reacts with water to form iron oxide and hydrogen gas. This is a redox reaction where electron exchange occurs. - In equation (b), iron sulfide reacts with oxygen to form iron oxide and sulfur dioxide. This is a combustion reaction involving oxygen. For both reactions, identifying the type of reaction is the first step. Recognizing whether it's a synthesis, decomposition, single replacement, or double replacement reaction provides valuable clues to the stoichiometry involved and helps in balancing the chemical equation efficiently.

Molecular Representation

Molecular representation involves illustrating chemical formulas in a way that details the elemental composition and structure of molecules involved in reactions. These representations are crucial for understanding and visualizing the changes that occur during chemical reactions.

For instance, in the equation \( 3 \ \mathrm{Fe} + 4 \ \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Fe}_{3}\mathrm{O}_{4} + 4 \ \mathrm{H}_{2} \), knowing the molecular formula of water, \( \mathrm{H}_2\mathrm{O} \), helps in identifying the hydrogen and oxygen atoms involved in the reaction. Similarly, in \( 4 \ \mathrm{FeS} + 7 \ \mathrm{O}_2 \rightarrow 2 \ \mathrm{Fe}_2\mathrm{O}_3 + 4 \ \mathrm{SO}_2 \), understanding that \( \mathrm{FeS} \) and \( \mathrm{SO}_2 \) imply specific bonding configurations aids in visualizing the reaction.

When you write down a reaction, it's crucial to represent each reactant and product with the correct chemical symbol and formula, as this ensures accurate counting and balancing. Molecular formulas show not only the amount but also ways these atoms are arranged, which directly affects the reaction's outcome. As a visualization tool, molecular representation is fundamental in studying and understanding chemical reactions.