Question

Balance each of the following chemical equations by inspection: (a) \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathrm{NH}_{3}(g)\) (b) \(\mathrm{Al}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}_{2}(g)\) (c) \(\operatorname{Sr}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \operatorname{Sr}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{KOH}(a q)\) (e) \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{Mn}(\mathrm{OH})_{2}(\mathrm{~s}) \rightarrow \mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) 3H2 + N2 → 2NH3 (b) Al2(CO3)3 → Al2O3 + 3CO2 (c) Sr + 2H2O → Sr(OH)2 + H2 (d) K2SO4 + Ba(OH)2 → BaSO4 + 2KOH (e) 2H3PO4 + 3Mn(OH)2 → Mn3(PO4)2 + 6H2O

Step-by-step solution

Balance equation (a) - Components Inventory

Identify the number of atoms of each element in the reactants and products. The reactants are \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\), with 2 hydrogen and 2 nitrogen atoms. The product is \(\mathrm{NH}_{3}(g)\) with 1 nitrogen and 3 hydrogen atoms.

Balance equation (a) - Equalizing Atoms

To balance nitrogen, place a coefficient of 2 before \(\mathrm{NH}_{3}\) to have 2 nitrogen atoms. This makes the equation: \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\). Now balance hydrogen by placing a 3 before \(\mathrm{H}_{2}(g)\): \(3\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\).

Balance equation (b) - Components Inventory

Start by identifying the elements and their counts: \(\mathrm{Al}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)\) has 2 Al, 3 C, and 9 O atoms. The products, \(\mathrm{Al}_{2}\mathrm{O}_{3}(s)\) and \(\mathrm{CO}_{2}(g)\), contain 2 Al and 3 C, 2 in \(\mathrm{CO}_{2}\).

Balance equation (b) - Equalizing Atoms

Since Aluminum and Carbon are already balanced, focus on Oxygen. There are 9 O in the reactant and only 6 (3 in \(\mathrm{Al}_{2}\mathrm{O}_{3}\) and 6 from 3 \(\mathrm{CO}_{2}\)), so 3 \(\mathrm{CO}_{2}(g)\) are enough. The balanced equation is: \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3}(s) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) + 3\mathrm{CO}_{2}(g)\).

Balance equation (c) - Components Inventory

Start by counting the elements: Sr(s), \(\mathrm{H}_{2}\mathrm{O}(l)\) with 2 H, 1 O; the products are \(\mathrm{Sr}(\mathrm{OH})_{2}(aq)\) and \(\mathrm{H}_{2}(g)\).

Balance equation (c) - Equalizing Atoms

To balance Sr, place 1 mol on each side. For hydrogen, put a 2 in front of \(\mathrm{H}_{2}\mathrm{O}\) to balance the 4 H atoms: \(\operatorname{Sr}(s)+2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \operatorname{Sr}(\mathrm{OH})_{2}(aq)+\mathrm{H}_{2}(g)\).

Balance equation (d) - Components Inventory

Inventory: \(\mathrm{K}_{2}\mathrm{SO}_{4}\) with 2 K, 1 S, 4 O in SO4; \(\mathrm{Ba}(\mathrm{OH})_{2}\) with 1 Ba, 2 O, and 2 H. Products have \(\mathrm{BaSO}_{4}\) and \(\mathrm{KOH}\) with 1 K, 1 O, and 1 H.

Balance equation (d) - Equalizing Atoms

Place a 2 in front of \(\mathrm{KOH}\) to balance 2 K and 2 OH from \(\mathrm{Ba}(\mathrm{OH})_{2}\). This balances all elements: \(\mathrm{K}_{2}\mathrm{SO}_{4}(aq) + \mathrm{Ba}(\mathrm{OH})_{2}(aq) \rightarrow \mathrm{BaSO}_{4}(s) + 2\mathrm{KOH}(aq)\).

Balance equation (e) - Components Inventory

Identify elements in reactants: \(\mathrm{H}_{3}\mathrm{PO}_{4}\) with 3 H, 1 P, 4 O; \(\mathrm{Mn}(\mathrm{OH})_{2}\) with 1 Mn, 2 O, 2 H. The products \(\mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}\) have 3 Mn, 6 O, 2 P, and \(\mathrm{H}_{2}\mathrm{O}\) with 2 H, 1 O.

Balance equation (e) - Equalizing Atoms

To balance Mn, put 3 in front of \(\mathrm{Mn}(\mathrm{OH})_{2}\). Next, 2 \(\mathrm{H}_{3}\mathrm{PO}_{4}\) for P and O: \(2\mathrm{H}_{3}\mathrm{PO}_{4}(aq)+3\mathrm{Mn}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Mn}_{3}(\mathrm{PO}_{4})_{2}(s)+{6}\mathrm{H}_{2}\mathrm{O}(l)\). This balances H with 6 on both sides.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Think of it as the mathematical part of chemistry, where you calculate the amounts of substances involved. When you balance a chemical equation, you are essentially applying the principles of stoichiometry to ensure that the number of atoms of each element is equal on both sides of the equation.

One of the keys to mastering stoichiometry is to understand coefficients in a chemical equation. Coefficients are the numbers placed before the chemical substances, indicating how many molecules or moles are involved in the reaction. For instance, in the equation \(3\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\), the coefficient '3' before \(\mathrm{H}_{2}\) suggests three moles of hydrogen react.

Stoichiometry serves as the backbone of quantitative chemistry, helping chemists predict the outcomes of reactions, determine yield, and scale reactions for industrial processes. Remember, accurate stoichiometric calculations require balanced chemical equations first, ensuring that the calculations reflect reality.

Reactants and Products

In every chemical reaction, we encounter reactants and products. Understanding the distinction between these two is crucial for chemical reactions.

• **Reactants:** These are the starting substances that undergo change during the reaction. They are typically found on the left-hand side of the chemical equation. For example, in the reaction \(\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathrm{NH}_{3}(g)\), the reactants are \(\mathrm{H}_{2}\) and \(\mathrm{N}_{2}\).
• **Products:** These are the substances that are formed as a result of the chemical reaction. They are found on the right-hand side of the equation. In our example, \(\mathrm{NH}_{3}(g)\) is the product formed.

Reactants are like raw materials that chemically transform during the reaction. In contrast, products are the new substances produced from these changes. This concept helps chemists understand how matter is conserved and rearranged in reactions. When balancing equations, it is critical to identify and count each atom in the reactants and products to ensure they comply with the laws of chemistry.

Chemical Reactions

Chemical reactions are the processes by which substances interact to form new compounds. An important aspect of chemical reactions is that they involve rearrangements of atoms, but they do not create or destroy atoms. This concept lies at the heart of all chemical transformations.

In a chemical reaction, bonds between atoms in the reactants break, and new bonds form to create the products. There are various types of chemical reactions, including:

• **Synthesis reactions,** where simpler substances combine to form a more complex product.
• **Decomposition reactions,** where a compound breaks down into simpler substances.
• **Replacement reactions,** where elements in different compounds swap places.

Each type of reaction involves specific patterns of breaking and forming bonds, and understanding these can help predict the products of a reaction. Chemical equations are the symbolic representation of these reactions. They allow scientists to convey all aspects of the chemical changes efficiently.

Law of Conservation of Mass

The Law of Conservation of Mass is a fundamental principle in chemistry that states that matter cannot be created or destroyed in a closed system. It is crucial when balancing chemical equations.

According to this law, the mass of the reactants must equal the mass of the products in a chemical reaction. This is why it is necessary to balance chemical equations with care; each side of the equation must have an equal number of each type of atom. For example, in the equation \(3\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\), the number of hydrogen and nitrogen atoms is the same on both sides, demonstrating mass balance.

The Law of Conservation of Mass is an integral part of understanding chemical processes and supports the idea that although substances change form during a reaction, the total amount of matter remains constant. This law is applicable in all chemical equations, and understanding it is essential for both theoretical and practical chemistry.