Problem 69 - Chemistry The Central Science In Si Units

Question

Liquefied petroleum gas (LPG) consists primarily of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(I)\) 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 \(0.50 \mathrm{~g} / \mathrm{mL}\)

Step-by-Step Solution

Answer

(a) \(\mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\); (b) \(2\mathrm{C}_{3}\mathrm{H}_{8} + 7\mathrm{O}_{2} \rightarrow 6\mathrm{CO} + 8\mathrm{H}_{2}\mathrm{O}\); (c) 60.27 L air.

Step 1: Balance Propane Combustion Equation

A complete combustion reaction of propane involves reacting it with oxygen to produce carbon dioxide and water. The unbalanced equation is: \[ \mathrm{C}_{3}\mathrm{H}_{8}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \] To balance the equation, we need equal amounts of each type of atom on both sides.Balanced equation: \[ \mathrm{C}_{3}\mathrm{H}_{8}(l) + 5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \]

Step 2: Balance Incomplete Combustion Equation

In incomplete combustion, propane reacts with less oxygen, resulting in carbon monoxide and water. The equation is: \[ \mathrm{C}_{3}\mathrm{H}_{8}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(l) \] After balancing, we get: \[ 2\mathrm{C}_{3}\mathrm{H}_{8}(l) + 7\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}(g) + 8\mathrm{H}_{2}\mathrm{O}(l) \]

Step 3: Calculate Propane Mass

To find the mass of propane, use the volume and density: \[ \text{mass} = \text{volume} \times \text{density} \] With 10 mL of propane and a density of 0.50 g/mL:\[ 10.0 \text{ mL} \times 0.50 \text{ g/mL} = 5.0 \text{ g} \]

Step 4: Convert Mass to Moles

Convert the mass of propane to moles using molar mass. The molar mass of \(\mathrm{C}_{3}\mathrm{H}_{8}\) is 44.1 g/mol. \[ \frac{5.0 \text{ g}}{44.1 \text{ g/mol}} = 0.113 \, \text{moles} \]

Step 5: Determine Oxygen Required for Combustion

From the balanced equation, 1 mole of \(\mathrm{C}_{3}\mathrm{H}_{8}\) requires 5 moles of \(\mathrm{O}_{2}\). For 0.113 moles of \(\mathrm{C}_{3}\mathrm{H}_{8}\):\[ 5 \times 0.113 = 0.565 \text{ moles of } \mathrm{O}_{2} \]

Step 6: Convert Moles to Volume of Oxygen

At STP, 1 mole of gas occupies 22.4 L. Therefore, the volume of \(\mathrm{O}_{2}\) needed is:\[ 0.565 \text{ moles} \times 22.4 \text{ L/mole} = 12.656 \text{ L} \]

Step 7: Calculate Air Volume Required for Combustion

Dry air is 21% oxygen by volume. Therefore, the total volume of air required is:\[ \frac{12.656 \text{ L}}{0.21} = 60.27 \text{ L} \] This is the minimum quantity of air needed.

Key Concepts

Propane Combustion

Propane is a versatile hydrocarbon that is commonly used as a fuel. It undergoes combustion reactions in the presence of oxygen, which is a chemical reaction where a substance combines with oxygen to produce heat. One of the key applications of propane is in the form of liquefied petroleum gas (LPG), primarily composed of propane, and sometimes butane.
In the complete combustion of propane, the chemical reaction involves propane reacting with oxygen to produce carbon dioxide and water. This process is highly exothermic, meaning it releases a significant amount of energy. The balanced chemical equation for the complete combustion of propane is: \[ \mathrm{C}_{3}\mathrm{H}_{8}(l) + 5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \]
This formula ensures that all carbon atoms from propane become carbon dioxide, and all hydrogen atoms become water.

Incomplete Combustion

Incomplete combustion of propane occurs when there is insufficient oxygen present during the reaction. In this case, instead of carbon dioxide, carbon monoxide and water are produced.
Carbon monoxide is a colorless and odorless gas that can be hazardous when inhaled in large quantities. The equation that represents incomplete combustion of propane is: \[ 2\mathrm{C}_{3}\mathrm{H}_{8}(l) + 7\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}(g) + 8\mathrm{H}_{2}\mathrm{O}(l) \]

• The limited supply of oxygen prevents the complete breakdown of carbon, leading to carbon monoxide.
• Incomplete combustion is less efficient and produces different exhaust products.

Proper ventilation is crucial in areas where combustion occurs to prevent the accumulation of dangerous gases like carbon monoxide.

Balanced Chemical Equation

A balanced chemical equation is essential for understanding how reactants transform into products. It ensures the law of conservation of mass, meaning that atoms are neither created nor destroyed in a chemical reaction.
To balance a chemical equation, adjust the coefficients of reactants and products so that the number of atoms of each element is the same on both sides of the equation.

• Start by balancing the most complex molecule first, often the hydrocarbon in combustion reactions.
• Next, balance the oxygen and hydrogen atoms.
• Finally, double-check to ensure each type of atom balances.

For the combustion of propane, the balanced equation ensures all atoms from the propane and oxygen are accounted for in the water and carbon dioxide produced.

Mole Calculations

In chemistry, mole calculations are crucial to relate the mass of substances involved in reactions to the volume of gases produced or consumed.
A mole is a unit that represents Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles, whether they are atoms, molecules, etc.
For example, when you convert mass to moles, you use the formula: \[ \text{moles} = \frac{\text{mass of substance}}{\text{molar mass}} \]

• This allows you to find out how much of a substance you have in terms of particles.
• Moles help to predict and calculate other substances involved in the reaction.

For the combustion of propane, using mole calculations, you can determine how much oxygen is required or how much carbon dioxide is produced based on how many moles of propane combust.