Question

Balance the following chemical equations. \(\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2} \mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)\) \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)\) \(\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

The balanced chemical equations are: 1. \(2\mathrm{MnO}_{2}(s) + 4\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q) + 2\mathrm{CO}_{2}(g)\) 2. \(2\mathrm{Al}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 3\mathrm{H}_{2}(g)\) 3. \(\mathrm{C}_{4}\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\) 4. \(2\mathrm{NH}_{4} \mathrm{I}(a q) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NH}_{4} \mathrm{Cl}(a q) + \mathrm{I}_{2}(g)\) 5. \(2\mathrm{KOH}(a q) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-step solution

Balancing Equation 1

\(\mathrm{MnO}_{2}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q)+\mathrm{CO}_{2}(g)\) 1. First, balance the manganese (\(\text{Mn}\)) atoms: \(2 \times \mathrm{MnO}_{2}\rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}\) 2. Then, balance the oxygen (\(\text{O}\)) atoms: \(4 \times \mathrm{CO}\rightarrow 2 \times \mathrm{CO}_{2}\) The balanced equation is: \(2\mathrm{MnO}_{2}(s) + 4\mathrm{CO}(g) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(a q) + 2\mathrm{CO}_{2}(g)\)

Balancing Equation 2

\(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) 1. Start with balancing the aluminium (\(\text{Al}\)) atoms: \(2 \times \mathrm{Al} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) 2. Then, balance the sulfate (\(\mathrm{SO}_{4}\)) ions: \(3 \times \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) 3. Finally, balance the hydrogen (\(\text{H}\)) atoms: \(3 \times \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 6 \times \mathrm{H}_{2}\) The balanced equation is: \(2\mathrm{Al}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q) + 3\mathrm{H}_{2}(g)$

Balancing Equation 3

\(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) 1. Start by balancing the carbon (\(\text{C}\)) atoms: \(\mathrm{C}_{4}\mathrm{H}_{10} \rightarrow 4 \times \mathrm{CO}_{2}\) 2. Then, balance the hydrogen (\(\text{H}\)) atoms: \(\mathrm{C}_{4}\mathrm{H}_{10} \rightarrow 5 \times \mathrm{H}_{2} \mathrm{O}\) 3. Finally, balance the oxygen (\(\text{O}\)) atoms: \(13/2 \times \mathrm{O}_{2} \rightarrow 4 \times \mathrm{CO}_{2} + 5 \times \mathrm{H}_{2} \mathrm{O}\) The balanced equation is: \(\mathrm{C}_{4}\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\)

Balancing Equation 4

\(\mathrm{NH}_{4} \mathrm{I}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)+\mathrm{I}_{2}(g)\) 1. First, balance the iodine (\(\text{I}\)) atoms: \(2 \times \mathrm{NH}_{4}\mathrm{I} \rightarrow \mathrm{I}_{2}\) 2. Then, balance the chlorine (\(\text{Cl}\)) atoms: \(\mathrm{NH}_{4}\mathrm{I} + \mathrm{Cl}_{2} \rightarrow 2 \times \mathrm{NH}_{4}\mathrm{Cl}\) The balanced equation is: \(2\mathrm{NH}_{4} \mathrm{I}(a q) + \mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{NH}_{4} \mathrm{Cl}(a q) + \mathrm{I}_{2}(g)\)

Balancing Equation 5

\(\mathrm{KOH}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) 1. First, balance the potassium (\(\text{K}\)) atoms: \(2 \times \mathrm{KOH} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4}\) 2. Then, balance the sulfate (\(\mathrm{SO}_{4}\)) ions: \(\mathrm{KOH}+ \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4}\) 3. Finally, balance the water (\(\mathrm{H}_{2}\mathrm{O}\)) molecules: \(2 \times \mathrm{KOH} \rightarrow 2 \times \mathrm{H}_{2}\mathrm{O}\) The balanced equation is: \(2\mathrm{KOH}(a q) + \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}(a q) + 2\mathrm{H}_{2} \mathrm{O}(l)\)

Key concepts

Stoichiometry

Stoichiometry can be thought of as the 'mathematics' behind chemistry. It provides quantitative relationships between substances involved in a chemical reaction. In practice, stoichiometry is used to calculate the amount of reactants required to produce a desired amount of product. A stoichiometric coefficient represents the number of molecules or moles of a substance involved in a reaction.

Let's consider an example: in the chemical equation \(2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}\), the stoichiometric coefficients are 2 for hydrogen, 1 for oxygen, and 2 for water. They indicate that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

When balancing equations, it’s crucial to adjust these coefficients to respect the conservation of mass, ensuring that the same number of atoms for each element is present on both sides of the equation. Understanding stoichiometry is not only fundamental for balancing chemical equations but also essential for empirical calculations, determining reaction yields, and performing conversions such as moles to grams, and vice versa.

Chemical Reaction

A chemical reaction is a process in which substances, known as reactants, transform into new substances called products. During this process, the bonds between atoms are broken and new ones are formed. There are different types of chemical reactions, including combination, decomposition, single-replacement, double-replacement, and combustion.

Each chemical reaction is represented by a chemical equation which shows the symbols and formulas of the reactants and products. For instance, in the combustion reaction \(\mathrm{C}_4\mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_2(g) + 5\mathrm{H}_2 \mathrm{O}(l)\), butane gas (\text{C}_4\text{H}_{10}) reacts with oxygen (\text{O}_2) to produce carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}).

For a chemical reaction to be useful and accurate, the equation representing it must be balanced so that the same number of each type of atom appears on both the reactant and product sides. This principle is known as the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of a substance to the number of particles it contains. A mole is defined as exactly 6.02214076 x 10^23 particles (Avogadro's number), which could be atoms, molecules, ions, or electrons depending on the context.

For example, one mole of carbon atoms weighs 12 grams, and because of Avogadro's number, it contains exactly 6.022 x 10^23 carbon atoms. The mole concept allows chemists to count atoms and molecules by weighing, which is a pragmatic approach to handling vast numbers of atomic or molecular entities.

Understanding the mole concept is crucial in stoichiometry for conversions between the mass of a substance and the amount in moles. It is critical to recognize that when we balance a chemical equation, we are actually balancing the moles of the reactants and products according to the ratios provided by the balanced equation, ensuring that mass and the number of particles are conserved.