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

11.36 liters of \(\text{CO}_2\) are produced at STP.

Step-by-step solution

Write the Chemical Equation

The combustion of propane \(\text{C}_3\text{H}_8\) in oxygen \(\text{O}_2\) to form carbon dioxide \(\text{CO}_2\) and water \(\text{H}_2\text{O}\) can be represented by:\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \] This is the balanced chemical equation for the reaction.

Calculate Moles of Propane

The molar mass of propane is \(3 \times 12.01 + 8 \times 1.01 = 44.11 \text{ g/mol}\). Calculate the moles of propane: \[ \text{Moles of C}_3\text{H}_8 = \frac{7.45 \text{ g}}{44.11 \text{ g/mol}} \approx 0.169 \text{ mol} \]

Use Stoichiometry to Find Moles of \(\text{CO}_2\)

From the balanced equation, 1 mole of \(\text{C}_3\text{H}_8\) produces 3 moles of \(\text{CO}_2\). Therefore, the moles of \(\text{CO}_2\) produced are:\[ 0.169 \text{ mol C}_3\text{H}_8 \times \frac{3 \text{ mol CO}_2}{1 \text{ mol C}_3\text{H}_8} = 0.507 \text{ mol CO}_2 \]

Convert Moles \(\text{CO}_2\) to Liters at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \(\text{CO}_2\) is:\[0.507 \text{ mol CO}_2 \times 22.4 \text{ L/mol} = 11.36 \text{ L CO}_2 \]

Key concepts

Understanding Stoichiometry

Stoichiometry is essentially the accountant of the chemical world, allowing us to keep track of quantities in reactions. It helps us determine the right amount of reactants to use and predict the amount of products formed. Imagine it as following a recipe in cooking. Just like you need specific amounts of ingredients for a dish, in chemistry, stoichiometry ensures that chemical reactions are well-balanced. In the propane combustion example, stoichiometry involves using a balanced equation, where we can see that 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. This balance allows us to use mole ratios to calculate how much product will form from a given amount of reactant.

Calculating Molar Mass

Molar mass is a key concept that helps us convert between mass and moles, which is essential for stoichiometric calculations. It is the mass of one mole of a substance, measured in grams per mole (g/mol). For example, the molar mass of propane \( \text{C}_3\text{H}_8 \), is calculated by adding the molar masses of its constituent elements: three carbon atoms and eight hydrogen atoms. Carbon has a molar mass of 12.01 g/mol and hydrogen has 1.01 g/mol. Thus, propane has a molar mass of \( 3 \times 12.01 + 8 \times 1.01 = 44.11 \) g/mol. This ratio facilitates the conversion from mass to moles, which is necessary to determine how much of a substance is involved in a reaction.

Standard Temperature and Pressure (STP) Conditions

Standard Temperature and Pressure, or STP, is a reference point used in chemistry to measure gases. It is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. At STP, one mole of any ideal gas occupies a volume of 22.4 liters. This provides a simple way to approximate the behavior of gases in various conditions, making calculations more manageable, especially in stoichiometry. In our propane combustion problem, knowing that 1 mole of carbon dioxide gas occupies 22.4 liters at STP, we can quickly convert the moles of \(\text{CO}_2\) obtained from combustion into liters, providing a tangible measurement of the gas produced.

The Process of Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance reacts quickly with oxygen, releasing energy in the form of light and heat. They are crucial in our daily lives, powering engines and heating homes. The generalized combustion reaction for a hydrocarbon, like propane \(\text{C}_3\text{H}_8\), involves the reactants being hydrocarbon and oxygen, and the products being carbon dioxide and water. The chemical equation for the combustion of propane is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]. This balanced equation indicates that for every one mole of propane combusted, five moles of oxygen are needed to fully react, resulting in the production of three moles of carbon dioxide and four moles of water. Understanding this allows us to accurately predict product amounts from given reactants, which is fundamental in optimizing energy outputs and resource usage.