Question

Complete and balance each combustion reaction equation. $${a}.\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \quad \text { b. } \mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow$$ $$ {c}. \mathrm{Ca}(s)+\mathrm{O}_{2}(g) \longrightarrow \quad \text { d. } \mathrm{C}_{5} \mathrm{H}_{12} \mathrm{S}(l)+\mathrm{O}_{2}(g) \longrightarrow $$

Short answer

a. \( \mathrm{S}(s) + \mathrm{O}_2(g) \longrightarrow \mathrm{SO}_2(g) \) b. \( \mathrm{C}_3\mathrm{H}_6(g) + \frac{9}{2}\mathrm{O}_2(g) \longrightarrow 3\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(g) \) c. \( 2\mathrm{Ca}(s) + \mathrm{O}_2(g) \longrightarrow 2\mathrm{CaO}(s) \) d. \( \mathrm{C}_5\mathrm{H}_{12}\mathrm{S}(l) + 13\mathrm{O}_2(g) \longrightarrow 5\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(g) + \mathrm{SO}_2(g) \)

Step-by-step solution

- Balance Sulphur Combustion

For the reaction \(\mathrm{S}(s) + \mathrm{O}_2(g) \longrightarrow \mathrm{SO}_2(g)\), start by balancing the sulphur atoms. There is one sulphur atom on both sides. Then, balance the oxygen atoms by adding a coefficient of 2 on the product side for \(\mathrm{O}_2(g)\), resulting in \(\mathrm{S}(s) + \mathrm{O}_2(g) \longrightarrow \mathrm{SO}_2(g)\).

- Balance Propane Combustion

For \(\mathrm{C}_3\mathrm{H}_6(g) + \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(g)\), balance carbon atoms first: 3 \(\mathrm{C}_3\mathrm{H}_6\) molecules to produce 3 \(\mathrm{CO}_2\). Then balance hydrogen, with 3 \(\mathrm{H}_2\mathrm{O}\) molecules. Finally, balance oxygen, resulting in 9/2 (or 4.5 if fractional coefficients are allowed) \(\mathrm{O}_2\) molecules on the reactant side for a balanced equation of \(\mathrm{C}_3\mathrm{H}_6(g) + \frac{9}{2}\mathrm{O}_2(g) \longrightarrow 3\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(g)\).

- Balance Calcium Combustion

For the reaction \(\mathrm{Ca}(s) + \mathrm{O}_2(g) \longrightarrow \mathrm{CaO}(s)\), start by balancing the calcium atoms. There is one calcium atom on both sides. Then balance the oxygen atoms. For every one molecule of \(\mathrm{O}_2\), there are two \(\mathrm{CaO}\) molecules produced. Therefore, the balanced equation is \(2\mathrm{Ca}(s) + \mathrm{O}_2(g) \longrightarrow 2\mathrm{CaO}(s)\).

- Balance Thiopentane Combustion

For \(\mathrm{C}_5\mathrm{H}_{12}\mathrm{S}(l) + \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(g) + \mathrm{SO}_2(g)\), start by balancing carbon atoms with 5 \(\mathrm{CO}_2\) molecules. Balance hydrogen with 6 \(\mathrm{H}_2\mathrm{O}\) molecules. The sulphur is already balanced. Finally, balance oxygen, which are 13 \(\mathrm{O}_2\) molecules on the reactant side, resulting in the balanced equation \(\mathrm{C}_5\mathrm{H}_{12}\mathrm{S}(l) + 13\mathrm{O}_2(g) \longrightarrow 5\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(g) + \mathrm{SO}_2(g)\).

Key concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the law of conservation of mass is upheld, meaning that atoms are neither created nor destroyed in a chemical reaction. To achieve a balanced equation, each side must have the same number of atoms of each element.

Take the combustion of sulfur, represented as \( \text{S}(s) + \text{O}_2(g) \longrightarrow \text{SO}_2(g) \). We must ensure that the number of each type of atom is equal on both sides. In this case, we have one sulfur atom on each side and two oxygen atoms (as part of one \text{O}_2 molecule) on the reactant side that produce two oxygen atoms in the product \text{SO}_2. The equation is already balanced. For a more complex compound like propane \( \text{C}_3\text{H}_6(g) \), the process involves balancing multiple elements, but the principle remains the same.

Understanding this concept is critical, not just for academic purposes, but also for real-world applications, such as calculating the required amount of reactants for industrial chemical processes.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between the substances involved in chemical reactions. It is an essential concept for calculating reactants and products in chemical equations.

Consider the propane combustion equation \( \text{C}_3\text{H}_6(g) + \text{O}_2(g) \longrightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \). Stoichiometry allows us to calculate how much \text{O}_2 is needed to combust a certain amount of \text{C}_3\text{H}_6, and how much \text{CO}_2 and \text{H}_2\text{O} will be produced. It is based on the coefficients of the balanced equation, which represent the ratios of molecules that react and are produced. Understanding this concept is crucial for students, as it forms the basis for many laboratory and industrial calculations.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are chemical processes involving the transfer of electrons between two species. An oxidation reaction involves the loss of electrons, while reduction involves the gain of electrons. Combustion reactions, like those in the given exercise, are a type of redox reaction where the fuel is oxidized by oxygen.

For instance, in the combustion of thiopentane \( \text{C}_5\text{H}_{12}\text{S}(l) + \text{O}_2(g) \longrightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) + \text{SO}_2(g) \), the sulphur in thiopentane is oxidized to \text{SO}_2, and carbon is oxidized to \text{CO}_2. Being able to identify oxidation and reduction components within a redox reaction is an important skill in chemistry. It not only aids in balancing complex reactions but also in understanding the electronic and energy changes that occur during chemical transformations.