Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from \(7.45 \mathrm{~g}\) of propane.

Short answer

The balanced equation for this reaction is \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\). The volume of carbon dioxide that could be produced from 7.45 g of propane is 11.36 L

Step-by-step solution

Write the unbalanced chemical equation

First, it is important to write the reaction of propane burning in oxygen, which results in carbon dioxide and water vapor. The initial, unbalanced, equation is: \(\mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)

Balance the chemical equation

Here, we will balance the equation by providing the required number of molecules for each reactant and product. The balanced equation is: \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\)

Calculate moles of propane

Before we can find the number of liters of \(\mathrm{CO}_{2}\), we need to find the number of moles of propane. This can be done using the formula: moles = mass / molar mass. Here, the molar mass of propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)) is 44.10 g/mol. Thus the number of moles of propane is \(7.45 g / 44.10 g/mol = 0.169 mol\)

Calculate moles of carbon dioxide

In the balanced chemical equation, it is clear that 1 mole of propane gives 3 moles of carbon dioxide. Therefore, the number of moles of \(\mathrm{CO}_{2}\) produced by \(0.169 mol\) of propane is \(0.169 mol * 3 = 0.507 mol\)

Calculate volume of carbon dioxide

Lastly, using the molar volume of a gas at STP which is 22.4L/mol, the volume of carbon dioxide can be calculated as \(0.507 mol * 22.4 L/mol = 11.36 L\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the conservation of mass. In any chemical reaction, the mass and atoms on the reactant side must equal those on the product side. This means that for every atom appearing in the reactants, there's an equivalent atom in the products.
To balance a chemical equation, follow these steps:

• Write the unbalanced equation, listing all reactants and products with their respective chemical formulas.
• Count the number of each type of atom in the reactants and products.
• Adjust coefficients in front of compounds to balance the number of atoms on each side of the equation.
• Ensure all coefficients are in the smallest possible whole-number ratio.

A practical example involves balancing the combustion of propane. Initially, you might write: \[\mathrm{C}_{3} \mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\]Upon balancing, each side has 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms, leading to:\[\mathrm{C}_{3} \mathrm{H}_{8} + 5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2}\mathrm{O}\] This balanced equation accounts for the conservation of atoms and illustrates proper stoichiometry.

Molar Volume of Gas

The molar volume of a gas is an essential concept in stoichiometry that connects volume measurements to mole calculations. At standard temperature and pressure (STP), which are defined as 0°C and 1 atm, one mole of any ideal gas occupies 22.4 liters. This is critical when converting between moles and volume in gaseous reactions.Imagine you want to know how much volume a certain amount of gas will occupy at STP. Simply multiply the number of moles by the molar volume (22.4 L/mol). For example, if you have 0.507 moles of carbon dioxide, its volume can be calculated as:\[0.507 \hspace{1mm} \text{moles} \times 22.4 \hspace{1mm} \text{L/mol} = 11.36 \hspace{1mm} \text{liters}\]This calculation demonstrates how the molar volume acts as a bridge between moles and volume, making it easier to quantify the gases produced or used in reactions. Understanding this relationship is key to solving problems involving gases.

Moles and Molar Mass

Understanding moles and molar mass is pivotal in explaining how substances react quantitatively. A mole is a unit that relates quantities of substances on a macroscopic scale to molecules or atoms on a microscopic scale. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).To find the number of moles from a given mass, use the formula:\[\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\]Consider propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)), which has a molar mass of 44.10 g/mol. If you have 7.45 g of propane, the moles are calculated by:\[\frac{7.45 \hspace{1mm} \text{g}}{44.10 \hspace{1mm} \text{g/mol}} = 0.169 \hspace{1mm} \text{moles}\]This method helps quantify reactants and products, determining how much of a substance participates or is generated in a chemical reaction. By interrelating mass, moles, and molar mass, stoichiometry becomes a practical tool for predicting the outcomes of reactions.

Combustion Reactions

Combustion reactions are a class of chemical reactions where a substance combines with oxygen, releasing energy in the form of heat and light. These reactions are essential in daily life, providing energy in various forms like burning fuels.A typical combustion reaction involves hydrocarbons, compounds made of hydrogen and carbon. When hydrocarbons like propane combust, they react with oxygen to produce carbon dioxide and water, represented by the balanced equation:\[\mathrm{C}_{3} \mathrm{H}_{8} + 5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2}\mathrm{O}\]In this reaction:

• Propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) acts as the fuel.
• Oxygen (\(\mathrm{O}_{2}\)) is the oxidizing agent.
• The products are carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)).

Combustion reactions are exothermic, meaning they release energy, making them useful in engines, heating, and electricity generation. By understanding these reactions, we can better design systems for energy management and reduce environmental impacts.