Question

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a 20.4 -gram sample of pentane is burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?

Short answer

When a 20.4-gram sample of pentane is burned in excess oxygen, approximately 30.57 grams of water can be produced, assuming a 100% yield.

Step-by-step solution

Balance the chemical equation

First, balance the chemical equation for the combustion of pentane: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) $$ The balanced chemical equation is: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) + 8\mathrm{O}_{2}(g) \rightarrow 5\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(l) $$

Convert mass of pentane to moles

Next, determine the moles of pentane (C5H12). Given mass of pentane is 20.4g. The molar mass of pentane can be calculated as follows: $$ \mathrm{C}_{5} \mathrm{H}_{12}: 5 \times 12.01\ \mathrm{g/mol\ (for\ C)} + 12 \times 1.01\ \mathrm{g/mol\ (for\ H)} = 72.15\ \mathrm{g/mol} $$ Now, we convert the mass of pentane to moles: $$ \mathrm{moles\ of\ pentane} = \frac{20.4\ \mathrm{g}}{72.15\ \mathrm{g/mol}} \approx 0.283\ \mathrm{mol} $$

Determine the stoichiometric ratio of pentane to water

Now, we find the stoichiometric ratio of pentane to water from the balanced chemical equation: $$ \frac{\mathrm{C}_{5} \mathrm{H}_{12}}{\mathrm{H}_{2} \mathrm{O}} = \frac{1}{6} $$

Calculate the moles of water produced

Using the stoichiometric ratio, determine the moles of water produced: $$ \mathrm{moles\ of\ water} = 0.283\ \mathrm{mol\ of\ pentane} \times \frac{6\ \mathrm{mol\ of\ water}}{1\ \mathrm{mol\ of\ pentane}} = 6 \times 0.283 = 1.698\ \mathrm{mol} $$

Convert moles of water to mass

Finally, determine the mass of water produced. The molar mass of water is: $$ \mathrm{H}_{2} \mathrm{O}: 2 \times 1.01\ \mathrm{g/mol\ (for\ H)} + 16.00\ \mathrm{g/mol\ (for\ O)} = 18.02\ \mathrm{g/mol} $$ Now, convert the moles of water to mass: $$ \mathrm{mass\ of\ water} = 1.698\ \mathrm{mol} \times 18.02\ \mathrm{g/mol} \approx 30.57\ \mathrm{g} $$ So, if a 20.4-gram sample of pentane is burned in excess oxygen, the mass of water produced is approximately 30.57 grams, assuming a 100% yield.

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the quantitative relationships between the substances as they participate in various chemical reactions. For students tackling problems involving chemical reactions, understanding stoichiometry is like having a roadmap: it allows them to navigate through the quantities of reactants and products involved.

In the exercise provided, stoichiometry comes into play when determining how much water is produced from the combustion of pentane. The balanced chemical equation serves as the basis for these stoichiometric calculations. It shows the exact ratio in which reactants combine and products form, which is crucial for predicting the outcome of the reaction in measurable terms. This is why the first step is always to ensure the equation is balanced, as the conservation of mass dictates that the number of atoms of each element must be the same on both the reactant and product sides of the equation.

Why is Stoichiometry Essential?

Stoichiometry is essential because it establishes the foundation for crucial concepts such as the conservation of mass and the mole ratio of reactants and products. It aids scientists and engineers in predicting yields of reactions and scaling them from laboratory to industrial scales. Without a balanced equation, these predictions would not be possible, and practical applications like the manufacturing of pharmaceuticals, materials, or fuel would be much harder to achieve efficiently and economically.

Mole Concept

The mole concept is a central pillar in the study of chemistry. It provides a bridge between the microscopic world of atoms and molecules and the macroscopic world that we can measure. One mole of a substance contains Avogadro's number of particles (atoms, molecules, or ions), which is approximately 6.022 x 1023 particles.

In our pentane combustion example, converting grams of pentane to moles allows us to use the balanced chemical equation to calculate the amount of water produced. This conversion is done using the molar mass of pentane, which is the sum of the atomic masses of the component atoms in a single molecule, expressed in grams per mole. This process clearly illustrates the practical application of the mole concept: it's how we scale up the properties of single molecules to quantities we can work with in the laboratory.

Practicality of the Mole Concept

The mole concept is not just an abstract idea; it's a tool that's instrumental for chemists when dealing with chemical reactions. For instance, in pharmacology, precise dosages of substances are crucial, which depend on the accurate use of the mole concept for their preparations. In the classroom, understanding the mole concept allows students to predict the outcomes of reactions and comprehend the significance behind the mass and volume measurements they work with.

Combustion Reaction

A combustion reaction is a type of chemical reaction that occurs when a substance combines with oxygen, releasing energy in the form of light or heat. In most cases, the products of a combustion reaction are carbon dioxide and water, especially when the fuel is a hydrocarbon, like pentane in the given exercise.

The combustion of pentane is highly exothermic, meaning it releases a significant amount of energy. These types of reactions are fundamental to various industries, such as energy production in power plants, engines in vehicles, and heating systems. In this particular combustion reaction, an understanding of the combustion process helps predict the amount of water produced—a question that could be relevant in environmental sciences and engineering where the by-products of fuel burning are of interest.

Real-world Application of Combustion Reactions

Examining combustion reactions like the burning of pentane provides insights into the development of cleaner burning fuels and improving energy efficiency. For students aspiring towards futures in renewable energy or environmental conservation, grappling with the combustion reaction equips them with the knowledge to tackle some of the pressing energy challenges of our times.