Question

Balance each of the following equations that describe combustion reactions. a. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

Balanced equations: a. \(C_{2}H_{6}(g) + \frac{7}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(g)\) b. \(C_{4}H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_{2}O(g)\) c. \(C_{6}H_{14}(l) + \frac{19}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 7H_{2}O(g)\)

Step-by-step solution

Balancing reaction a

To balance the reaction: (1) \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) 1. Balance the carbon atoms: There are 2 carbon atoms on the left-hand side and only 1 on the right-hand side. To balance, we'll put a coefficient of 2 in front of CO₂: (2) \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) 2. Balance the hydrogen atoms: There are 6 hydrogen atoms on the left-hand side and only 2 on the right-hand side. To balance, we'll put a coefficient of 3 in front of H₂O: (3) \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\) 3. Balance the oxygen atoms: There are now 7 oxygen atoms on the right-hand side and only 2 on the left-hand side. To balance, we'll put a coefficient of 7/2 in front of O₂: (4) \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\) Now, the reaction is balanced.

Balancing reaction b

To balance the reaction: (1) \(C_{4} H_{10}(g)+O_{2}(g) \rightarrow CO_{2}(g)+H_{2} O(g)\) 1. Balance the carbon atoms: There are 4 carbon atoms on the left-hand side and only 1 on the right-hand side. To balance, we'll put a coefficient of 4 in front of CO₂: (2) \(C_{4} H_{10}(g)+O_{2}(g) \rightarrow 4 CO_{2}(g)+H_{2} O(g)\) 2. Balance the hydrogen atoms: There are 10 hydrogen atoms on the left-hand side and only 2 on the right-hand side. To balance, we'll put a coefficient of 5 in front of H₂O: (3) \(C_{4} H_{10}(g)+O_{2}(g) \rightarrow 4 CO_{2}(g)+5 H_{2} O(g)\) 3. Balance the oxygen atoms: There are now 13 oxygen atoms on the right-hand side and only 2 on the left-hand side. To balance, we'll put a coefficient of 13/2 in front of O₂: (4) \(C_{4} H_{10}(g)+\frac{13}{2} O_{2}(g) \rightarrow 4 CO_{2}(g)+5 H_{2} O(g)\) Now, the reaction is balanced.

Balancing reaction c

To balance the reaction: (1) \(C_{6} H_{14}(l)+O_{2}(g) \rightarrow CO_{2}(g)+H_{2} O(g)\) 1. Balance the carbon atoms: There are 6 carbon atoms on the left-hand side and only 1 on the right-hand side. To balance, we'll put a coefficient of 6 in front of CO₂: (2) \(C_{6} H_{14}(l)+O_{2}(g) \rightarrow 6 CO_{2}(g)+H_{2} O(g)\) 2. Balance the hydrogen atoms: There are 14 hydrogen atoms on the left-hand side and only 2 on the right-hand side. To balance, we'll put a coefficient of 7 in front of H₂O: (3) \(C_{6} H_{14}(l)+O_{2}(g) \rightarrow 6 CO_{2}(g)+7 H_{2} O(g)\) 3. Balance the oxygen atoms: There are now 19 oxygen atoms on the right-hand side and only 2 on the left-hand side. To balance, we'll put a coefficient of 19/2 in front of O₂: (4) \(C_{6} H_{14}(l)+\frac{19}{2} O_{2}(g) \rightarrow 6 CO_{2}(g)+7 H_{2} O(g)\) Now, the reaction is balanced.

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Essentially, it's like a recipe for chemistry, where the ingredients (reactants) must be measured precisely to create a certain amount of product.

To master stoichiometry, one must be comfortable with the mole concept, which allows chemists to count particles using a unit called the 'mole', much like one would count eggs by the dozen. A mole corresponds to Avogadro’s number of particles, which is approximately 6.022 x 10²³ particles.

When balancing combustion reactions, stoichiometry involves determining the right proportions of reactants and products to ensure that atoms are conserved during the reaction. Correct stoichiometry is crucial in applications ranging from cooking to industrial chemical synthesis, where the correct product yield is paramount.

Chemical Equation Balancing

The art of balancing chemical equations involves ensuring that the same number of each type of atom is present on both sides of the equation. This conservation of mass mirrors real-life reactions where atoms are neither created nor destroyed, merely rearranged.

When balancing a combustion reaction, which is a type of chemical reaction where a substance combines with oxygen to produce heat and light, one must often start by balancing the carbon and hydrogen atoms before tackling the more complex oxygen atoms, as seen in the provided step-by-step solutions.

Detailed Breakdown of the Process

First, balance the carbons by adjusting coefficients in front of the carbon dioxide, CO_2. Next, balance hydrogens by tweaking the coefficients for water, H_2O. Finally, check and balance the oxygens, which may require using fractional coefficients for the O_2 molecule, understanding that in practice, chemical equations are represented with whole numbers.

Combustion Reaction

A combustion reaction is a high-energy reaction that occurs when a fuel and an oxidant (typically oxygen) react, producing heat and often light. Common fuels include hydrocarbons, substances made of carbon and hydrogen, such as in the textbook exercises.

These reactions are exothermic, meaning they release energy, and they form the basis for many everyday applications like vehicle engines and power plants. The general form of a combustion reaction involves a hydrocarbon reacting with oxygen to produce carbon dioxide and water. Balancing these reactions is not only important for chemistry homework but also for practical applications such as calculating the energy output of a fuel or the required amount of oxygen for complete combustion.

Mole Concept

The mole concept is a fundamental idea in chemistry that enables scientists to speak the same language when discussing the number of atoms, ions, or molecules in a sample. The mole is based on Avogadro's number and provides a bridge between the microscopic world of atoms and the macroscopic world we observe.

Understanding the mole concept is essential when performing stoichiometric calculations in chemical reactions. It allows for the translation from moles to grams via molar mass and then to the observable quantities measured in a laboratory or used in industrial processes. A firm grasp of the mole concept is the first step towards accurately predicting product amounts from given reactants or determining the proportions required for a reaction to occur precisely as intended.