Question

Write a balanced equation for the complete combustion for each of the following: (a) ethane (b) butane (c) pentane (d) heptane

Short answer

(a) 2C_2H_6 + 7O_2 → 4CO_2 + 6H_2O; (b) 2C_4H_{10} + 13O_2 → 8CO_2 + 10H_2O; (c) C_5H_{12} + 8O_2 → 5CO_2 + 6H_2O; (d) C_7H_{16} + 11O_2 → 7CO_2 + 8H_2O.

Step-by-step solution

Identify the General Formula for Combustion

The complete combustion of a hydrocarbon involves reacting it with oxygen to produce carbon dioxide and water. The general formula for the combustion of a hydrocarbon \\( C_xH_y \) is: \\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \] In order to balance the equation, we need to balance the number of carbon, hydrogen, and oxygen atoms on both sides of the equation.

Write the Chemical Formula for Each Hydrocarbon

Before writing the combustion equations, we need to know the chemical formulas for the hydrocarbons: \ (a) Ethane: \( C_2H_6 \) \ (b) Butane: \( C_4H_{10} \) \ (c) Pentane: \( C_5H_{12} \) \ (d) Heptane: \( C_7H_{16} \)

Balance the Combustion Equation for Ethane

Using the formula for ethane \( C_2H_6 \), the combustion equation before balancing is: \\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \] To balance it: \- Balance carbon: Place 2 in front of \( CO_2 \), giving \( C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \)- Balance hydrogen: Place 3 in front of \( H_2O \), giving \( C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \)- Balance oxygen: We have 7 oxygen atoms on the right, so place 3.5 or 7/2 in front of \( O_2 \): \( C_2H_6 + \frac{7}{2} O_2 \rightarrow 2CO_2 + 3H_2O \)- To get integer coefficients, multiply the entire equation by 2: \\[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \]

Balance the Combustion Equation for Butane

For butane \( C_4H_{10} \), the combustion equation is: \\[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \] Balance it as follows:- Balance carbon: Place 4 in front of \( CO_2 \), giving \( C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O \)- Balance hydrogen: Place 5 in front of \( H_2O \), giving \( C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O \)- Balance oxygen: We have 13 oxygen atoms on the right, so place 6.5 or 13/2 in front of \( O_2 \): \( C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4CO_2 + 5H_2O \)- Multiply through by 2 to clear fractions: \\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]

Balance the Combustion Equation for Pentane

For pentane \( C_5H_{12} \), the combustion equation is: \\[ C_5H_{12} + O_2 \rightarrow CO_2 + H_2O \] Balance as: - Carbon: Place 5 in front of \( CO_2 \), giving \( C_5H_{12} + O_2 \rightarrow 5CO_2 + H_2O \)- Hydrogen: Place 6 in front of \( H_2O \), giving \( C_5H_{12} + O_2 \rightarrow 5CO_2 + 6H_2O \)- Oxygen: We have 16 oxygens on the right, so place 8 in front of \( O_2 \): \\[ C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O \]

Balance the Combustion Equation for Heptane

For heptane \( C_7H_{16} \), the combustion is: \\[ C_7H_{16} + O_2 \rightarrow CO_2 + H_2O \] Balance as: - Carbon: Place 7 in front of \( CO_2 \), giving \( C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O \)- Hydrogen: Place 8 in front of \( H_2O \), giving \( C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O \)- Oxygen: We have 22 oxygens on the right, so place 11 in front of \( O_2 \): \\[ C_7H_{16} + 11O_2 \rightarrow 7CO_2 + 8H_2O \]

Key concepts

Combustion Reactions

Combustion reactions are a vital type of chemical reaction involving oxygen and a fuel, typically a hydrocarbon. These reactions are exothermic, meaning they release energy—primarily in the form of heat and light. In a complete combustion reaction of a hydrocarbon, the products are typically carbon dioxide and water. This conversion requires sufficient oxygen so that all the carbon atoms in the fuel form carbon dioxide, and all the hydrogen atoms form water molecules.

• The basic idea of combustion is simple: a hydrocarbon fuel reacts with oxygen.
• Complete combustion results in carbon dioxide and water as the sole products.
• Balancing these equations ensures that the number of each type of atom on both sides of the equation is equal.

For students, balancing combustion reactions often involves guessing and checking the coefficients in front of molecules to make sure the number of atoms matches on each side. It can be a puzzle, but understanding the basic concept that every carbon produces a carbon dioxide and every hydrogen forms water helps immensely.

Hydrocarbons

Hydrocarbons are compounds made up of hydrogen and carbon atoms only. They are the primary fuels in combustion reactions and are found in gasoline, natural gases, and various oils. Hydrocarbons are classified into different types such as alkanes, alkenes, and alkynes based on the types of bonds between the carbon atoms:

• Alkanes, like ethane, butane, pentane, and heptane, are saturated hydrocarbons. They contain single covalent bonds between carbon atoms.
• Each type of hydrocarbon has a general chemical formula: Alkanes follow the formula \(C_nH_{2n+2}\).
• Hydrocarbons are used extensively as fuels because they release a high amount of energy during combustion.

Understanding the properties of hydrocarbons, particularly their structure and how they react, is crucial for mastering combustion reactions. By learning the basic formulas, students can easily write and balance combustion equations for any given hydrocarbon.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative aspects of chemical reactions. It involves calculations based on balanced chemical equations, allowing us to predict the amounts of substances consumed and produced in reactions.

• Balanced equations are essential as they ensure the mass and the number of atoms are conserved.
• Stoichiometry helps in determining the proportions of reactants and products in a chemical reaction.
• When balancing chemical equations, consider the conservation of mass and the law of definite proportions—which states that a chemical compound always contains the exact proportion of elements by mass.

By mastering stoichiometry, students can understand and predict the outcomes of reactions with precision. It's particularly important in combustion reactions where the exact amount of oxygen is required to ensure complete combustion. Balancing equations like those for ethane, butane, pentane, and heptane relies heavily on these stoichiometric principles.