Question

Balance each of the following chemical equations: a. \(\mathrm{Al}+\mathrm{O}_{2} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}\) b. \(\mathrm{NO}+\mathrm{O}_{2} \rightarrow \mathrm{NO}_{2}\) c. \(\mathrm{H}_{2}+\mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Fe}+\mathrm{H}_{2} \mathrm{O}\)

Short answer

a. \(4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\) b. \(2\mathrm{NO} + \mathrm{O}_2 \rightarrow 2\mathrm{NO}_2\) c. \(2\mathrm{H}_2 + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow 2\mathrm{Fe} + 3\mathrm{H}_2\mathrm{O}\)

Step-by-step solution

Balance Al and O for reaction a

Start by balancing the aluminum (Al) atoms. Since there are 2 Al atoms in \(\mathrm{Al}_2\mathrm{O}_3\), put a coefficient of 2 in front of \(\mathrm{Al}\) on the reactant side: \(2\mathrm{Al} + \mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3\). Now, balance the oxygen (O) atoms. There are 3 O atoms in \(\mathrm{Al}_2\mathrm{O}_3\), so put a coefficient of \(\frac{3}{2}\) in front of \(\mathrm{O}_2\) to get 3 O atoms: \(2\mathrm{Al} + \frac{3}{2}\mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3\). To avoid fractions, multiply all coefficients by 2: \(4\mathrm{Al} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\).

Balance NO and O for reaction b

Balance the nitrogen (N) atoms first by placing a coefficient of 2 in front of \(\mathrm{NO}\) since there are 2 N atoms in \(2\mathrm{NO}_2\): \(2\mathrm{NO} + \mathrm{O}_2 \rightarrow 2\mathrm{NO}_2\). Now, balance the oxygen atoms. There are a total of 4 O atoms on the product side (in \(2\mathrm{NO}_2\)), and you already have 2 O atoms from \(\mathrm{O}_2\), so the equation is already balanced.

Balance H2, Fe2O3, Fe, and H2O for reaction c

Start with \(\mathrm{Fe}_2\mathrm{O}_3\) since it contains more than one element. Keep \(\mathrm{Fe}_2\mathrm{O}_3\) with a coefficient of 1. There are 2 Fe atoms on the reactant side, so place a coefficient of 2 in front of Fe on the product side: \(\mathrm{H}_2 + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow 2\mathrm{Fe} + \mathrm{H}_2\mathrm{O}\). Now, balance the hydrogen (H) atoms. Because there is 1 \(\mathrm{H}_2\mathrm{O}\) molecule on the product side with 2 H atoms, you need to place a coefficient of 2 in front of \(\mathrm{H}_2\) on the reactant side: \(2\mathrm{H}_2 + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow 2\mathrm{Fe} + \mathrm{H}_2\mathrm{O}\). Lastly, balance the oxygen (O) atoms. There are 3 O atoms in \(\mathrm{Fe}_2\mathrm{O}_3\), so place a coefficient of 3 in front of \(\mathrm{H}_2\mathrm{O}\): \(2\mathrm{H}_2 + \mathrm{Fe}_2\mathrm{O}_3 \rightarrow 2\mathrm{Fe} + 3\mathrm{H}_2\mathrm{O}\). The equation is now balanced.

Key concepts

Stoichiometry

Stoichiometry is similar to a recipe in cooking—it outlines the amounts of reactants (ingredients) needed to create a certain number of products (dishes). But instead of cups and teaspoons, stoichiometry uses 'moles', a unit to express amounts of a chemical substance.

In chemistry, it is not enough to know which substances react with each other; we must also know the proportions at which they react. This is determined through coefficients placed in front of the chemical formulas in a balanced chemical equation, ensuring that the number of atoms of each element is the same on both the reactant and product sides. For example, in the equation balancing exercise shown above, stoichiometry explains why we need exactly four moles of aluminum to react with three moles of oxygen to produce two moles of aluminum oxide.

Understanding stoichiometry is crucial because it allows chemists to predict the outcomes of reactions, determine the amounts of substances needed for a reaction, and calculate yields—the amount of product obtained from a reaction.

Chemical Reaction

At the heart of stoichiometry lies the chemical reaction, a process where substances, known as reactants, are transformed into different substances, called products. During a chemical reaction, the bonds between atoms in the reactants break, and new bonds form to create the products.

In the given exercise, we see examples of different types of chemical reactions. Reaction (a) is a synthesis reaction, where elements combine to form a compound. On the other hand, reaction (c) represents a single displacement reaction, where an element displaces another in a compound, forming a new element and compound. When understanding reactions, it's important not just to balance them, but also to understand the context in which these reactions occur, such as the temperature, pressure, and presence of catalysts, which can affect the reaction's rate and yield.

Law of Conservation of Mass

The Law of Conservation of Mass is a fundamental principle that states that mass cannot be created or destroyed in a closed system through ordinary chemical reactions or physical changes. It means that the mass of the reactants must equal the mass of the products.

In the context of balancing chemical equations, this law requires that the same number of each type of atom is present after the chemical reaction as before. The exercise above demonstrates this principle through the careful balancing of atoms on both sides of the equation. No atoms are lost or gained during the reactions; they are simply rearranged. This concept reinforces why stoichiometry is necessary: to ensure that matter is preserved, following the Law of Conservation of Mass, we must use stoichiometrically correct proportions of reactants to form the desired products.