Question

Balance each of the following equations. Classify each reaction as synthesis, decomposition, single-displacement, or doubledisplacement. (a) \(\mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{NaOH} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Al}_{2}\left(\mathrm{CO}_{3}\right)_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{CO}_{2}\) (e) \(\mathrm{NH}_{4} \mathrm{I}+\mathrm{Cl}_{2} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{I}_{2}\)

Short answer

Balanced reactions: (a) \(2 \mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) - Synthesis, (b) \(\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow 2 \mathrm{NH}_{3} + \mathrm{N}_{2}\) - Decomposition, (c) \(\mathrm{H}_{2} \mathrm{SO}_{4} + 2\mathrm{NaOH} \longrightarrow 2\mathrm{H}_{2} \mathrm{O} + \mathrm{Na}_{2} \mathrm{SO}_{4}\) - Double-displacement, (d) \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3} + 3\mathrm{CO}_{2}\) - Decomposition, (e) \(2 \mathrm{NH}_{4} \mathrm{I} + \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NH}_{4} \mathrm{Cl} + \mathrm{I}_{2}\) - Single-displacement.

Step-by-step solution

Balancing the Equation (a)

Inspect the atoms of each element on both sides of the equation. For \( \mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \), balance oxygen by adding a coefficient of 2 to \(\mathrm{H}_{2} \mathrm{O}\). The balanced equation is \(2 \mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\). Classify the reaction as a synthesis reaction, where two or more simple substances combine to form a more complex substance.

Balancing the Equation (b)

Write down the unbalanced equation \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{NH}_{3}(g) + \mathrm{N}_{2}(g)\). To balance nitrogen, we already have the correct number of nitrogen atoms. The hydrogen atoms are not balanced. We need two moles of ammonia to balance the hydrogen atoms, making the balanced equation \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 2 \mathrm{NH}_{3}(g) + \mathrm{N}_{2}(g)\). This reaction is a decomposition reaction, where a single compound breaks down into two or more simpler substances.

Balancing the Equation (c)

Examine the atoms on both sides for the reaction \(\mathrm{H}_{2} \mathrm{SO}_{4} + \mathrm{NaOH} \longrightarrow \mathrm{H}_{2} \mathrm{O} + \mathrm{Na}_{2} \mathrm{SO}_{4}\). To balance sodium and sulfate ions, add a coefficient of 2 in front of \(\mathrm{NaOH}\) and \(\mathrm{H}_{2} \mathrm{O}\). The balanced equation becomes \(\mathrm{H}_{2} \mathrm{SO}_{4} + 2\mathrm{NaOH} \longrightarrow 2\mathrm{H}_{2} \mathrm{O} + \mathrm{Na}_{2} \mathrm{SO}_{4}\). This is a double-displacement reaction where two ionic compounds exchange ions to form two new compounds.

Balancing the Equation (d)

Focus on balancing the atoms for \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3} + \mathrm{CO}_{2}\). Add a coefficient of 3 to \(\mathrm{CO}_{2}\) to balance carbon atoms. The corresponding balanced equation is \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3} + 3\mathrm{CO}_{2}\). This is also a decomposition reaction, as heating \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3}\) results in it breaking down into simpler substances.

Balancing the Equation (e)

Balance the equation \(\mathrm{NH}_{4} \mathrm{I} + \mathrm{Cl}_{2} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl} + \mathrm{I}_{2}\). By adding a coefficient of 2 to both \(\mathrm{NH}_{4} \mathrm{I}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\), we balance the reaction as \(2 \mathrm{NH}_{4} \mathrm{I} + \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NH}_{4} \mathrm{Cl} + \mathrm{I}_{2}\). This is a single-displacement reaction where one element (chlorine) replaces another element (iodine) in a compound.

Key concepts

Synthesis Reaction

In chemistry, a synthesis reaction is one of the most fundamental types of reactions. It involves the combination of two or more simple substances to create a more complex compound. An easy way to recognize a synthesis reaction is by its general formula: \( A + B \longrightarrow AB \).

In the provided exercise, equation (a) \(2 \mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) exemplifies a synthesis reaction. Here, two dihydrogen molecules (\(\mathrm{H}_{2}\)) react with one dioxygen molecule (\(\mathrm{O}_{2}\)) to yield two water molecules (\(\mathrm{H}_{2} \mathrm{O}\)). The reaction is balanced by ensuring an equal number of atoms of each element on both sides of the equation.

Key Pointers for Synthesis Reactions:

• Simple substances combine to form a single more complex product.
• Energy is typically released (exothermic) in these reactions.
• They are vital in the formation of complex molecules necessary for various biological and industrial processes.

Decomposition Reaction

A decomposition reaction is the opposite of a synthesis reaction. It involves a single compound breaking down into two or more simpler substances. The general formula for a decomposition reaction is \( AB \longrightarrow A + B \).

Exercise parts (b) and (d) demonstrate decomposition reactions. In equation (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow 2 \mathrm{NH}_{3}(g) + \mathrm{N}_{2}(g)\), the compound hydrazine decomposes into ammonia and nitrogen gas. In equation (d) \(\mathrm{Al}_{2}(\mathrm{CO}_{3})_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3} + 3\mathrm{CO}_{2}\), aluminum carbonate breaks down into aluminum oxide and carbon dioxide upon heating (indicated by the triangle above the arrow,\(\Delta\)).

Key Pointers for Decomposition Reactions:

• Require energy input, often as heat, to break chemical bonds.
• Involved in processes such as electrolysis, digestion, and the breakdown of organic matter.
• Products are generally simpler and can include elements and/or simpler compounds.

Single-Displacement Reaction

A single-displacement reaction involves an element and a compound reacting to form a new element and a new compound. One element is replaced by another in a compound. The general form can be represented as \( A + BC \longrightarrow B + AC \) where element A replaces B in the compound BC.

In the exercise, equation (e) \(2 \mathrm{NH}_{4} \mathrm{I} + \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NH}_{4} \mathrm{Cl} + \mathrm{I}_{2}\) is a single-displacement reaction. Chlorine (\(\mathrm{Cl}_{2}\)) displaces iodine (I) in ammonium iodide (\(\mathrm{NH}_{4} \mathrm{I}\)) to form ammonium chloride (\(\mathrm{NH}_{4} \mathrm{Cl}\)) and iodine (\(\mathrm{I}_{2}\)).

Key Pointers for Single-Displacement Reactions:

• Involve an exchange of an atom or ion between a molecule and a free element.
• Common in redox reactions where oxidation and reduction occur.
• Useful in extracting metals from ores and in galvanic cells.

Double-Displacement Reaction

Double-displacement reactions are characterized by the exchange of ions between two reactant ionic compounds. They generally take the form \( AB + CD \longrightarrow AD + CB \), where the cations and anions switch places.

Equation (c) from the exercise, \(\mathrm{H}_{2} \mathrm{SO}_{4} + 2\mathrm{NaOH} \longrightarrow 2\mathrm{H}_{2} \mathrm{O} + \mathrm{Na}_{2} \mathrm{SO}_{4}\), is a classic example of a double-displacement reaction, also known as a metathesis reaction. Here, the hydrogen ion (H) in sulphuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) and the sodium ion (Na) in sodium hydroxide (\(\mathrm{NaOH}\)) exchange places, forming water (\(\mathrm{H}_{2} \mathrm{O}\)) and sodium sulfate (\(\mathrm{Na}_{2} \mathrm{SO}_{4}\)).

Key Pointers for Double-Displacement Reactions:

• Typically occur in aqueous solutions where the reactants are dissolved in water.
• Often result in the formation of a precipitate, gas, or water.
• Frequently used in neutralization reactions where an acid reacts with a base to produce salt and water.