Question

Propane reacts with oxygen to give carbon dioxide and water vapor. $$ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If you mix \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio, and if the total pressure of the mixture is \(288 \mathrm{mm} \mathrm{Hg}\), what are the partial pressures of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{O}_{2} ?\) If the temperature and volume do not change, what is the pressure of the water vapor after reaction?

Short answer

Partial pressures: \(\mathrm{C}_{3}\mathrm{H}_{8}\) is 48 mm Hg, \(\mathrm{O}_{2}\) is 240 mm Hg; Water vapor is 192 mm Hg.

Step-by-step solution

Write the Balanced Equation

The balanced chemical equation for the combustion of propane is: \[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}) + 5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g}) + 4 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \] This equation 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 vapor.

Determine Mole Ratio

From the balanced equation, the mole ratio of \(\mathrm{C}_{3}\mathrm{H}_{8}\) to \(\mathrm{O}_{2}\) is 1:5. This means that for every one mole (or molecule) of propane, there are five moles (or molecules) of oxygen required for complete combustion.

Determine Partial Pressures from Total Pressure

If the mixture is in the correct stoichiometric ratio, then the partial pressures will reflect the same ratio. Let \(P_{\mathrm{C}_{3}\mathrm{H}_{8}}\) be the partial pressure of propane and \(P_{\mathrm{O}_{2}}\) be the partial pressure of oxygen. Thus, \(P_{\mathrm{C}_{3}\mathrm{H}_{8}} + P_{\mathrm{O}_{2}} = 288 \, \mathrm{mm} \, \mathrm{Hg}\) and \(5P_{\mathrm{C}_{3}\mathrm{H}_{8}} = P_{\mathrm{O}_{2}}\).

Solve for Partial Pressures

Using \(5P_{\mathrm{C}_{3}\mathrm{H}_{8}} = P_{\mathrm{O}_{2}}\), substitute into the total pressure equation: \(P_{\mathrm{C}_{3}\mathrm{H}_{8}} + 5P_{\mathrm{C}_{3}\mathrm{H}_{8}} = 288\). This simplifies to \(6P_{\mathrm{C}_{3}\mathrm{H}_{8}} = 288\), giving \(P_{\mathrm{C}_{3}\mathrm{H}_{8}} = 48\, \mathrm{mm} \, \mathrm{Hg}\). Using the earlier proportion, \(P_{\mathrm{O}_{2}} = 5 \times 48 = 240 \, \mathrm{mm} \, \mathrm{Hg}\).

Calculate Pressure of Water Vapor

Each mole of propane produces four moles of water vapor, according to the balanced equation. The pressure contributed by \(\mathrm{H}_{2}\mathrm{O}\) at the same conditions (temperature and volume constant) can be derived from its stoichiometric proportion relative to propane. \(P_{\mathrm{H}_{2}\mathrm{O}}\) equals \(4 \times P_{\mathrm{C}_{3}\mathrm{H}_{8}}\), so \(P_{\mathrm{H}_{2}\mathrm{O}} = 4 \times 48 = 192 \, \mathrm{mm} \, \mathrm{Hg}\).

Key concepts

Partial Pressures

When dealing with gases in a mixture, it's vital to understand the concept of partial pressures. Partial pressure is simply the pressure contributed by a single type of gas in a mixture of gases. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas.
In combustion reactions like the one involving propane and oxygen, if the gases are mixed in the correct stoichiometric amounts, their partial pressures will align with their mole ratios. For example, if propane (\( \mathrm{C}_{3} \mathrm{H}_{8} \)) and oxygen (\( \mathrm{O}_{2} \)) are mixed in a 1:5 mole ratio, then their partial pressures should also maintain this ratio.
This relationship helps us determine that, when the total pressure of the mixture is 288 mm Hg, the partial pressure of propane is 48 mm Hg while that of oxygen is 240 mm Hg.

Combustion Reaction

A combustion reaction is a vigorous chemical process that typically involves the burning of a fuel in an oxidizing agent, such as oxygen, resulting in the release of energy in the form of heat and light.
In the scenario of propane combusting in oxygen, a complete combustion occurs when propane fully reacts with oxygen to form water vapor and carbon dioxide.
Propane combustion can be represented by the equation: \[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}) + 5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g}) + 4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]This equation clearly shows the substances involved, their proportions, and the products formed during the reaction. During the process, propane acts as the fuel and oxygen as the oxidizer.

Balanced Chemical Equation

A balanced chemical equation is essential for correctly describing the quantities of reactants and products in a chemical reaction. This ensures that the law of conservation of mass is obeyed, meaning the total mass of reactants equals the total mass of products.
For the combustion of propane, the balanced chemical equation is:\[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}) + 5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g}) + 4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]This equation needs balancing because it must have the same number of each type of atom on both sides of the arrow. Here, it shows that 1 mole of propane reacts with 5 moles of oxygen to yield 3 moles of carbon dioxide and 4 moles of water vapor, fulfilling the balance of atoms for carbon, hydrogen, and oxygen.

Mole Ratio

Mole ratio is a key concept derived from the coefficients of a balanced chemical equation. It indicates the proportions of moles of reactants and products involved in the reaction. Understanding mole ratios allows chemists to predict how much of each reactant is needed to form a given amount of product.
In the combustion of propane, the balanced equation provides the mole ratio: for every 1 mole of \( \mathrm{C}_{3} \mathrm{H}_{8} \), there are 5 moles of \( \mathrm{O}_{2} \) required, 3 moles of \( \mathrm{CO}_{2} \) produced, and 4 moles of \( \mathrm{H}_{2} \mathrm{O} \) formed as the products.
This ratio of 1:5:3:4 is essential for calculating partial pressures and determining the correct proportions of gases involved in the reaction, ensuring that all reactants are consumed efficiently without any left over.