Question

Suppose that \(2.00 \mathrm{~L}\) of propane gas, \(\mathrm{C}_{3} \mathrm{H}_{8}\), at \(1.00\) atm and \(298 \mathrm{~K}\) is mixed with \(5.00 \mathrm{~L}\) of oxygen gas at the same pressure and temperature and burned to form carbon dioxide gas and liquid water. Ignore the volume of water formed, and determine the final volume of the reaction mixture (including products and excess reactant) at \(1.00 \mathrm{~atm}\) and \(298 \mathrm{~K}\) if reaction goes to completion.

Short answer

The final volume of the reaction mixture at STP is approximately \(3.00\mathrm{L}\) of \(\mathrm{CO}_2\) gas.

Step-by-step solution

Write the Balanced Chemical Equation

To solve the problem, first write the balanced chemical equation for the combustion of propane gas, \(\mathrm{C}_3\mathrm{H}_8\). The balanced equation is \(\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}\).

Determine the Limiting Reactant

Using the ideal gas law, convert the volumes of the gases into moles at STP (standard temperature and pressure) where \(1\mathrm{L}\) of gas equals \(1/22.4\mathrm{mol}\) at \(298\mathrm{K}\) and \(1.00\mathrm{atm}\). Propane has \(2.00\mathrm{L} \times (1/22.4\mathrm{mol/L}) = 0.0893\mathrm{mol}\) of \(\mathrm{C}_3\mathrm{H}_8\), and oxygen has \(5.00\mathrm{L} \times (1/22.4\mathrm{mol/L}) = 0.2232\mathrm{mol}\) of \(\mathrm{O}_2\). \(\mathrm{O}_2\) is the limiting reactant because it would be completely consumed first (since 5 moles of oxygen reacts with 1 mole of propane); \(0.2232\mathrm{mol} \times (1/5)=0.04464\mathrm{mol}\) of \(\mathrm{C}_3\mathrm{H}_8\) will react.

Calculate the Volume of Products

From the balanced equation, burning \(0.04464\mathrm{mol}\) of propane produces \(0.04464 \times 3 = 0.1339\mathrm{mol}\) of \(\mathrm{CO}_2\) and \(0.04464 \times 4 = 0.1786\mathrm{mol}\) of \(\mathrm{H}_2\mathrm{O}\) (liquid). The \(\mathrm{H}_2\mathrm{O}\) volume can be ignored. The leftover \(\mathrm{O}_2\) after reaction will be \(0.2232 - 0.2232 = 0\mathrm{mol}\), thus there is no excess \(\mathrm{O}_2\). The total volume of gas at STP is the volume of \(\mathrm{CO}_2\) which is \(0.1339\mathrm{mol} \times 22.4\mathrm{L/mol}=\;\)approximately \(3.00\mathrm{L}\).

Key concepts

Chemical Equations

Chemical equations are essential representations in chemistry that show the reactants and products involved in chemical reactions. Like any language, chemical equations use symbols and formulas to communicate a lot of information in a compact form. For example, the balanced chemical equation for the combustion of propane, \( \mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \), tells us that one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water. This equation is balanced, meaning the number of atoms for each element is the same on both sides of the equation, indicating the law of conservation of mass.
Understanding how to read and write these equations is crucial for predicting the outcomes of chemical reactions and for performing calculations involving the quantities of reactants and products.

Limiting Reactant

The concept of the limiting reactant is vital in stoichiometry, as it determines the amount of product that can be formed in a chemical reaction. It's the reactant that will be completely used up first, stopping the reaction from continuing, even if other reactants are still present. Identifying the limiting reactant involves comparing the mole ratio of the reactants used in the reaction to the mole ratio in the balanced chemical equation.
In our propane combustion example, oxygen was identified as the limiting reactant through a comparison of moles available and the stoichiometric requirements. Once you determine the limiting reactant, you can predict how much of the other reactants will be left over and how much product will be produced, ensuring efficient use of materials and aiding in the planning of chemical processes.

Ideal Gas Law

The ideal gas law is a foundational equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas via the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. While real gases deviate from ideal behavior under high pressure or low temperature, the ideal gas law is incredibly useful for solving problems under standard conditions, as seen in our propane combustion problem.
By using this law, we were able to convert volumes of propane and oxygen gases to moles, which is an essential step in calculations involving gaseous reactants and products. This concept is not limited to chemical reactions but also has wide applications in fields like engineering and meteorology.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the substances involved in chemical reactions. It helps us predict the amounts of reactants needed and products formed in a chemical reaction, based on the mole ratio provided by the balanced chemical equation. For instance, in the combustion of propane, stoichiometry allows us to calculate that 5 moles of oxygen react with 1 mole of propane to produce 3 moles of carbon dioxide and 4 moles of water, as dictated by the balanced equation.

Understanding stoichiometry is essential for anyone looking to work in the field of chemistry, as it is crucial for creating the desired products in the right amounts, which can impact cost-efficiency and environmental impact in industrial processes. It also demands a clear understanding of mole concept, molar mass, and Avogadro's number.