Question

Liquefied petroleum gas (LPG) consists primarily of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(l)\) or butane \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) (a) Write a balanced chemical equation for the complete combustion of propane to produce \(\mathrm{CO}_{2}(g)\) as the only carbon-containing product. (b) Write a balanced chemical equation for the incomplete combustion of propane to produce \(\mathrm{CO}(g)\) as the only carbon-containing product. (c) At \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) pressure, what is the minimum quantity of dry air needed to combust \(10.0 \mathrm{~mL}\) of \(\mathrm{C}_{3} \mathrm{H}_{8}(l)\) completely to \(\mathrm{CO}_{2}(g)\) ? The density of the LPG is \(0.50 \mathrm{~g} / \mathrm{mL}\).

Short answer

In short, for the complete combustion of propane, the balanced equation is: \[C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\] For the incomplete combustion of propane, the balanced equation is: \[2C_3H_8 + 7O_2 \rightarrow 4CO_2 + 2CO + 8H_2O\] Given 10.0 mL of propane with a density of 0.50 g/mL, the minimum quantity of dry air needed to combust it completely to CO2 gas at 25°C and 101.3 kPa is 0.05914 m³.

Step-by-step solution

Balanced Equation for Complete Combustion of Propane

To balance the complete combustion of propane, we need to produce CO2 and H2O as products. The equation is: \[C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\]

Balanced Equation for Incomplete Combustion of Propane

To balance the incomplete combustion of propane, we need to produce CO and H2O as products. The equation is: \[2C_3H_8 + 7O_2 \rightarrow 4CO_2 + 2CO + 8H_2O\]

Calculate the Mass of the Propane Sample

We are given that 10.0 mL of propane has a density of 0.50 g/mL. To find the mass of the propane sample, we multiply the volume by the density: Mass of propane = Volume × Density Mass of propane = 10.0 mL × 0.50 g/mL = 5.0 g

Moles of Propane

To find the moles of propane, we divide the mass by the molar mass (the molecular weight of propane, C3H8, is 44.10 g/mol): Moles of propane = Mass of propane / Molar mass of propane Moles of propane = 5.0 g / 44.10 g/mol = 0.1134 mol

Moles of O2 Required for Complete Combustion

Using stoichiometry, we can find the moles of O2 required for complete combustion. From the balanced equation for complete combustion, 1 mole of propane requires 5 moles of O2: Moles of O2 required = Moles of propane × (5 moles O2 / 1 mole propane) Moles of O2 required = 0.1134 mol × 5 = 0.5670 mol

Find the Volume of Air Needed

Dry air is about 21% oxygen by volume. We can use the ideal gas law (PV = nRT) to find the volume of oxygen required at the given conditions (T = 25°C = 298 K, P = 101.3 kPa), and then divide by the percentage of oxygen in the air to find the volume of air needed. R (ideal gas constant) = 8.314 kJ/(k . mol) Volume of O2 = n O2 × R × T / P Volume of O2 = 0.5670 mol × 8.314 kJ/(k . mol) × 298 K / 101.3 kPa = 0.01242 m³ Now, we need to find the volume of air needed: Volume of air = Volume of O2 / 0.21 (since air is 21% O2) Volume of air = 0.01242 m³ / 0.21 = 0.05914 m³ So, the minimum quantity of dry air needed to combust 10.0 mL of propane completely to CO2 gas is 0.05914 m³.

Key concepts

Combustion Reactions

Combustion reactions are chemical processes where a substance combines with oxygen, releasing heat and usually producing a flame. These reactions are crucial, especially in the context of everyday fuels like propane, a primary component of liquefied petroleum gas (LPG). Propane itself undergoes two types of combustion reactions: complete and incomplete.Complete Combustion
In complete combustion, propane combusts in the presence of an adequate oxygen supply. The chemical equation for complete combustion of propane is:\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \] This reaction produces carbon dioxide (CO₂) and water (H₂O) as products. Complete combustion is desirable in practical applications as it maximizes energy release and minimizes pollution.Incomplete Combustion
Incomplete combustion occurs when there is insufficient oxygen for the fuel to react completely. In this case, carbon monoxide (CO) and sometimes even carbon soot can form alongside water. The equation for the incomplete combustion of propane is:\[ 2C_3H_8 + 7O_2 \rightarrow 4CO_2 + 2CO + 8H_2O \] This reaction is less efficient, releasing less energy and producing harmful byproducts like carbon monoxide, which is a misconception people often encounter about combustion.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's key to predicting how much of each substance will be involved or produced in a chemical change, such as the combustion of propane.In the balanced equation for the complete combustion of propane, stoichiometry tells us that 1 mole of propane reacts with 5 moles of oxygen. Using this relationship, you can calculate the needed amounts of reactants to ensure the reaction proceeds completely.Finding Moles from Mass
Given a mass of propane, you can determine the number of moles using its molar mass. For instance:- The mass of propane given is 5.0 grams.- The molar mass of propane (C₃H₈) is 44.10 g/mol.- The moles of propane can be calculated as: \[ \text{Moles of propane} = \frac{5.0 \text{ g}}{44.10 \text{ g/mol}} = 0.1134 \text{ mol} \]Using Mole Ratios
Stoichiometry allows us to use mole ratios from the balanced equation to determine moles of other substances required or produced.- From complete combustion, 1 mole of propane requires 5 moles of O₂.- Hence, 0.1134 mol of propane would require: \[ 0.1134 \text{ mol} \times 5 = 0.5670 \text{ mol of O}_2 \]Understanding these relationships helps in calculating other related quantities, such as the volume of gases involved as achieved through stoichiometry.

Ideal Gas Law

The ideal gas law is a fundamental equation used to relate the pressure, volume, temperature, and moles of a gas. It is expressed as:\[ PV = nRT \]where:- \( P \) is the pressure in kilopascals (kPa),- \( V \) is the volume in cubic meters (m³),- \( n \) is the number of moles,- \( R \) is the ideal gas constant (8.314 kJ/(K·mol)),- \( T \) is the temperature in Kelvin (K).Applying the Ideal Gas Law
While working with gases like oxygen needed for the combustion of propane, this law helps determine the volume of oxygen required based on the moles calculated through stoichiometry at a given temperature and pressure.For instance, with the moles of O₂ needed being 0.5670 mol:- At a temperature of 25°C (or 298 K) and a pressure of 101.3 kPa:- The volume of O₂ can be calculated as: \[ V = \frac{nRT}{P} = \frac{0.5670 \times 8.314 \times 298}{101.3} = 0.01242 \text{ m}^3 \]Calculating Air Volume
To combust the propane completely, we must account for the fact that air is only 21% oxygen. Therefore:- The volume of air needed would be about: \[ \text{Volume of air} = \frac{0.01242}{0.21} = 0.05914 \text{ m}^3 \]Understanding and applying the ideal gas law is crucial for quantifying how much reactant is needed and ensuring combustion processes occur effectively under varying conditions.