Question

Complete and balance each of the following chemical equations that represent combustion reactions. a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow\) b. \(\mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow\) c. \(C_{7} H_{16}(l)+O_{2}(g) \rightarrow\)

Short answer

The balanced chemical equations for the combustion reactions are: a. \(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\) b. \(C_5H_{12}(l) + 8O_2(g) \rightarrow 5CO_2(g) + 6H_2O(l)\) c. \(C_7H_{16}(l) + 11O_2(g) \rightarrow 7CO_2(g) + 8H_2O(l)\)

Step-by-step solution

a. Balancing the combustion of propane (C3H8)

We start by listing the number of atoms of each element on both sides: C: 3 atoms on left (C3H8), 0 atoms on right H: 8 atoms on left (C3H8), 0 atoms on right O: 2 atoms on left (O2), 0 atoms on right Let's balance the equation: C3H8(g) + O2(g) -> 3CO2(g) + 4H2O(l) Now we have: C: 3 atoms on left, 3 atoms on right H: 8 atoms on left, 8 atoms on right O: 2 atoms on left, 10 atoms on right (6 in 3CO2 + 4 in 4H2O) We adjust the oxygen on the left side of the equation to balance the total oxygen atoms: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)

b. Balancing the combustion of pentane (C5H12)

We start by listing the number of atoms of each element on both sides: C: 5 atoms on left (C5H12), 0 atoms on right H: 12 atoms on left (C5H12), 0 atoms on right O: 2 atoms on left (O2), 0 atoms on right Let's balance the equation: C5H12(l) + O2(g) -> 5CO2(g) + 6H2O(l) Now we have: C: 5 atoms on left, 5 atoms on right H: 12 atoms on left, 12 atoms on right O: 2 atoms on left, 16 atoms on right (10 in 5CO2 + 6 in 6H2O) We adjust the oxygen on the left side of the equation to balance the total oxygen atoms: C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(l)

c. Balancing the combustion of heptane (C7H16)

We start by listing the number of atoms of each element on both sides: C: 7 atoms on left (C7H16), 0 atoms on right H: 16 atoms on left (C7H16), 0 atoms on right O: 2 atoms on left (O2), 0 atoms on right Let's balance the equation: C7H16(l) + O2(g) -> 7CO2(g) + 8H2O(l) Now we have: C: 7 atoms on left, 7 atoms on right H: 16 atoms on left, 16 atoms on right O: 2 atoms on left, 22 atoms on right (14 in 7CO2 + 8 in 8H2O) We adjust the oxygen on the left side of the equation to balance the total oxygen atoms: C7H16(l) + 11O2(g) -> 7CO2(g) + 8H2O(l)

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. Often, this substance is a hydrocarbon—a compound made up of carbon and hydrogen atoms—such as the propane (C3H8), pentane (C5H12), and heptane (C7H16) mentioned in the exercise. These reactions are exothermic, which means they emit energy. They're a pivotal concept in chemistry as they relate to everyday processes like burning fuels, cooking, and even cellular respiration.

The general formula for the combustion of a hydrocarbon is:
Hydrocarbon + Oxygen → Carbon Dioxide + Water

In a complete combustion reaction, the products are carbon dioxide (CO2) and water (H2O). If there's insufficient oxygen, incomplete combustion occurs, resulting in carbon monoxide (CO) and possibly soot (C) alongside water.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It's essentially the 'math of chemistry,' allowing scientists to predict the amounts of substances consumed and produced in a reaction. When solving the provided exercise, stoichiometry helps in determining the mole ratio between the reactants and products based on the coefficients in the balanced chemical equation.

Knowing the mole ratio is crucial because it shows the proportional relationship that dictates how much of each substance is needed or produced. For example, in the balanced reaction of propane provided, the stoichiometry indicates that one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.

Chemical Equation Balancing

Balancing chemical equations is an essential task in stoichiometry. It ensures that the law of conservation of mass is obeyed—matter is neither created nor destroyed in a chemical reaction. The number of atoms for each element must be the same on both the reactant and product sides of the equation.

Let's revisit the propane combustion example. Initially, we have three carbon atoms and eight hydrogen atoms. In a balanced equation, we need to ensure that these atoms are accounted for on the product side, hence the production of three carbon dioxide molecules and four water molecules.

However, balancing also requires the oxygen atoms to be equal on both sides. Since there are ten oxygen atoms on the product side (three from carbon dioxide and four from water), we have to adjust the reactant side to have ten oxygen atoms as well, which comes from five O2 molecules.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, colloquially known as redox reactions, involve the transfer of electrons between chemical species. Oxidation is the loss of electrons, and reduction is the gain of electrons. Combustion reactions are classic examples of redox reactions because the fuel substance (such as a hydrocarbon) is oxidized by oxygen.

In our combustion examples, the hydrocarbons are oxidized, meaning they lose electrons to the oxygen: the carbon in the hydrocarbons forms carbon dioxide, and the hydrogen forms water. Conversely, the oxygen molecules are reduced because they gain electrons from the hydrocarbons. Balancing redox equations can be more complex, as it must take into account both the mass and charge balance.